Circles

L OCUS L OCUS L OCUS L OCUS L OCUS 1 Mathematics / Circles Circles As we know very well from pure geometry, a circle is a geometrical figure described by a moving point in the Euclidean plane such that its distance from a fixed point is always constant. The fixed point is called the centre of the circle while the fixed distance is called the radius of the circle. Moving over to a co-ordinate system, let us denote the centre C of the circle by (x 0 , y 0 ) and the radius by r. For any point P(x, y) lying on the circle, the length PC must be equal to r. Using the distance formula for PC, we therefore obtain: ( ) ( ) 2 2 2 0 0 : Equation of thecircle x x y y r − + − · This equation must be satisfied by every point P(x, y) lying on the circle; therefore, this is the equation that uniquely describes the given circle. We simply call it the equation of the circle, with centre (x 0 , y 0 ) and radius r. For example, consider the circle with its centre at (1, 1) and radius equal to unity: Fig - 01 (1,1) y A circle with centre (11) and 1 ,r = x The equation of the this circle is ( ) ( ) 2 2 2 1 1 1 x y − + − · 2 2 2 2 1 0 x y x y ⇒ + − − + · Find the equation of the circle passing through the points (0, 0) , (3, 0) and (1, 2). Sec t i on-1 EQUATI ONS DESCRI BI NG CI RCLES Example – 01 L OCUS L OCUS L OCUS L OCUS L OCUS 2 Mathematics / Circles Solution: To write the equation of the required circle, we must find its centre and radius. Recall from pure geometry that a circle can always be drawn through three non-collinear points. This can be done as follows: join the points to form a triangle. Draw the perpendicular bisectors of any of the two sides of this triangle. Their point of intersection gives us the centre C. The distance of C from any of the vertices gives the radius r of the circles: C is the centre of the circle passing Fig - 02 C P Q R through P, Q, R. CP = CQ= CR = r. We apply this result to the current example: The equation of the perpendicular bisector of OB is 3 2 x · The equation of the perpendicular bisector y x O (0,0) B (3,0) C A (1, 2) Fig - 03 of OA is 1 1 1 2 2 y x − · − − 1 2 2 2 y x ⇒ − · − 5 2 0 2 x y ⇒ + − · The point C is the intersection of the two angle bisectors 3 1 , 2 2 C ¸ _ ⇒ ≡ ¸ , We can now easily evaluate the radius r as the length OC: 9 1 10 4 4 2 r OC · · + · Finally, the equation of the required circle becomes: 2 2 3 1 10 2 2 4 x y ¸ _ ¸ _ − + − · ¸ , ¸ , We will subsequently see another method to solve this type of questions. L OCUS L OCUS L OCUS L OCUS L OCUS 3 Mathematics / Circles Find the equation of the circle which touches the co-ordinate axes and whose centre lies on the line2 3 x y − · Solution: A circle of radius r touching the co-ordinate axes can be in one of the four following configurations, with four corresponding equations mentioned alongside: Fig - 04 ( ,) r r y x ( –) +xr 2 ( –)=yr r 2 2 (–,) r r y x ( ) +x + r 2 ( –)=yr r 2 2 ( ) r – r y x ( –) +xr 2 ( )=y + r r 2 2 (–,) r r – y x ( ) +x + r 2 ( )=y + r r 2 2 Note from these four possible cases that the centre of such a circle either lies on y = x or on y = – x. In the current example, the centre is also given to lie on x – 2y = 3. Thus, there will be two circles, with the two centres being given by the point of intersection y = x and y = – x with x – 2y =3. ( ) 1 and 2 3 3, 3 y x x y C · − · ⇒ ≡ − ⇒ Equation of the circle is ( ) ( ) 2 2 3 3 9 x y + + − · ( ) 2 and 2 3 1, 1 y x x y C · − − · ⇒ ≡ − ⇒ Equation of the circle is ( ) ( ) 2 2 1 1 1 x y − + + · Find the equation of the circle with radius 5 and which touches another circle 2 2 2 4 20 0 x y x y + − − − ·externally at the point (5, 5) Solution: Let us first try to rearrange the equation of the given circle in the standard form from which we’ll be able to deduce its centre and radius: 2 2 2 4 20 0 x y x y + − − − · ( ) ( ) 2 2 1 2 25 x y ⇒ − + − · Example – 02 Example – 03 L OCUS L OCUS L OCUS L OCUS L OCUS 4 Mathematics / Circles Therefore, the centre of this circle is (1, 2) and its radius is 5. We should now draw a geometrical figure which will certainly make things more clear: Fig - 05 y x () x, y 0 0 (5, 5) (1, 2) 5 The required circle Let the centre of the required circle be ( ,). Observe that is the mid-pt of (1,2) and ( ,) x y P x y 0 0 0 0 P The given circle As explained in the figure, we can now evaluate (x 0 , y 0 ), the centre of the required circle: 0 0 1 5 9 2 x x + · ⇒ · 0 0 2 5 8 2 y y + · ⇒ · Thus, the required equation is ( ) ( ) 2 2 9 8 25 x y − + − · EQUATION OF A CIRCLE : GENERAL FORM Expanding the standard form of the equation of the circle we derived in the last section, we’ll obtain: 2 2 2 2 2 0 0 0 0 2 2 0 x y x x yy x y r + − − + + − · This suggests that the most general form of the equation of a circle can be written in terms of three variables; call them g, f and c so that 2 2 2 0 0 0 0 2 2 ; 2 2 ; g x f y c x y r · − · − · + − 2 2 2 2 2 0 0 0 0 ; ; x g y f r x y c g f c ⇒ · − · − · + − · + − Thus, the equation of the circle in terms of g, f and c becomes 2 2 2 2 0 x y gx fy c + + + + · : Equation of a circle; most general form ( ) 2 2 Centre , ; Radius g f g f c · − − · + − It should be apparent to you how the standard form and the general form of the circle’s equation are interconvertible. Which form to use where is a matter of convenience and will depend on the situation. L OCUS L OCUS L OCUS L OCUS L OCUS 5 Mathematics / Circles As a first example, let us redo Example - 1, which involves finding the equation of the circle passing through the points (0, 0), (3, 0) and (1, 2). Let the equation be 2 2 2 2 0 x y gx fy c + + + + · , where g, f and c are to be determined. This equation must be satisfied by the three points through which the circle passes, and hence we’ll obtain three equations from which g, f and c can be determined: Substitute (0, 0) : c = 0 Substitute (3, 0) : 9 + 6g = 0 3 2 g ⇒ · − Substitute (1, 2) : 1 + 4 – 3 + 4f = 0 1 2 f ⇒ · − The required equation is hence: 2 2 3 0 x y x y + − − · which is the same as what we obtained in Example - 1. Let C be any circle with centre ( ) 0, 2 .Prove that at the most two rational points can lie on C. Solution: By a rational point, we mean a point which has both its co-ordinates rational. Let the equation of C be 2 2 2 2 0 x y gx fy c + + + + · We can arrive at the result easily be contradiction. Suppose that we have three rational points on the circle with the co-ordinates (x i , y i ) i = 1, 2, 3. These three points must satisfy the equation of the circle. Thus we obtain a system of linear equations in g and f : ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 2 2 2 2 0 2 2 0 2 2 2 2 0 2 2 x g y f c x y x y gx fy c x y gx fy c x g y f c x y x y gx fy c x g y f c x y ¹ ¹ + + · − + + + + + · ⇒ ¹ ¹ ¹ ¹ + + + + · ⇒ + + · − + ' ; ¹ ¹ + + + + · ⇒ + + · − + ¹ ¹ ¹ ¹ The coefficients in this system of linear equations are all rational by assumption. Thus, when we solve this system, we must obtain g, f and c to be all rational. But since the centre is ( ) 0, 2 , we have 2 f · − which gives us a contradiction. This means that our assumption of taking three rational points on the circle is wrong⇒ At the most two rational points can lie on thiscircle. Example – 04 L OCUS L OCUS L OCUS L OCUS L OCUS 6 Mathematics / Circles Suppose we are given two curves C 1 and C 2 whose equation are as follows: 2 2 1 1 1 1 1 1 1 : 2 2 2 0 C a x h xy b y g x f y c + + + + + · 2 2 2 2 2 2 2 2 2 : 2 2 2 0 C a x h xy b y gx f y c + + + + + · It is also given that these curves intersect in four concyclic points. Prove that 1 1 2 2 1 2 a b a b h h − − · Solution: From the discussions in the last chapter, we know that any curve C passing through the point(s) of intersection of two given curves 1 2 0 and 0 C C · ·can be written as 1 2 0 where C C C ≡ + λ · λ∈! We can do the same in the current example to obtain the equation of the curve passing through the four (concyclic) points of intersection as: ( ) ( ) ( ) ( ) ( ) 2 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 0 a a x h h xy b b y g g x f f y c c +λ + +λ + +λ + +λ + +λ + +λ · From the general form of the equation of the circle, we know that this equation (above) will represent the equation of a circle only if: Coeff. of x 2 = Coeff. of y 2 1 2 1 2 a a b b ⇒ + λ · + λ 1 1 2 2 b a b a − ⇒ λ · − − ... (1) Coeff. of xy = 0 1 2 0 h h ⇒ + λ · 1 2 h h ⇒ λ · − ... (2) From (1) and (2), we have 1 1 2 2 1 2 b a b a h h − − · Suppose that the equation of a circle is 2 2 2 2 0 S x y gx fy c ≡ + + + + · What condition must the co-ordinates of a point P(x 1 , y 1 ) satisfy so that P may lie (i) inside the circle (ii) outside the circle? Example – 05 Example – 06 L OCUS L OCUS L OCUS L OCUS L OCUS 7 Mathematics / Circles Solution: Let the centre of S be C and its radius be r. The point P lies inside Sif CP < r and outside S is CP > r. From the equation of S, we know C to be (–g, – f) and r to be 2 2 . g f c + − Using these facts, we can easily evaluate the required conditions: P lies inside the circle: 2 2 CP r < ( ) ( ) 2 2 2 2 1 1 x g y f g f c + + + < + − 2 2 1 1 1 1 2 2 0 x y gx fy c + + + + < ... (1) P lies outside the circle: 2 2 CP r > ( ) ( ) 2 2 2 2 1 1 x g y f g f c + + + > + − 2 2 1 1 1 1 2 2 0 x y gx fy c + + + + > ... (2) We can write (1) and (2) concisely as ( ) ( ) ( ) 1 1 1 1 1 1 lies inside the circle , 0 lies on the circle , 0 lies outside the circle , 0 P S x y P S x y P S x y ⇒ < ⇒ · ⇒ > Find the equation of the circle circumscribing the triangle formed by the lines6, 2 4 and 2 5 x y x y x y + · + · + · . Solution: Let us first consider the general case wherein we’ve been given three lines L 1 , L 2 and L 3 and we need to find the circle circumscribing the triangle that these three lines form Fig - 06 O C B A L 3 L 2 L 1 Example – 07 L OCUS L OCUS L OCUS L OCUS L OCUS 8 Mathematics / Circles One way to do it would be as follows: ! Find the intersection points A, B and C of the three lines ! Use these intersection points to write any two perpendicular bisectors ! Find the intersection of these two perpendicular bisectors which gives us the centre O. ! Finally, find the radius (which will equal OA, OB and OC) This procedure will definitely become quite lengthy. We look instead for a a more elegant method. We first try to write the equation of an arbitrary second-degree curve S passing through the intersection points of L 1 , L 2 and L 3 . Think carefully and you’ll realise that such a curve can be written in terms of two arbitrary constants λ and µ as follows: 1 2 2 3 3 1 0 S L L L L L L ≡ + λ +µ · That such a curve S will pass through all the three intersection points can be verified by observing that the substitution of the co-ordinates of any of the three points in the equation above will make both sides identically 0. One we have such a curve, we can impose the necessary constraints to make it a circle. Coming back to the current example, the equation of an arbitrary curve passing through the intersection points of the three lines can be written as: ( )( ) ( )( ) ( )( ) 6 2 4 2 4 2 5 2 5 6 0 S x y x y x y x y x y x y ≡ + − + − +λ + − + − +µ + − + − · To make S the equation of a circle, we simple impose the following constraints: 2 2 2 1 2 2 · ⇒ + λ +µ · + λ + µ 2 Coeff. of Coeff. of x y 1 ⇒ µ · .. (1) 3 5 3 0 ⇒ + λ + µ · Coeff. of = 0 xy 6 5 ⇒ λ · − ... (2) We substitute λ and µ back into S to obtain the required equation as: 2 2 17 19 50 0 S x y x y ≡ + − − + · As another example of following such an approach, suppose that we are given four straight lines and are told that they intersect at four concyclic points, as shown below: D L 2 Fig - 07 L 4 L 1 L 3 C B A A, B, C, D are four concyclic points L OCUS L OCUS L OCUS L OCUS L OCUS 9 Mathematics / Circles What approach will you follow if you’re told to find the equation of the circle circumscribing this quadrilateral? Obviously, one can always proceed by explicitly determining the centre and the radius of the said circle, but as in the previous question, a much more elegant method exists. Convince yourself that any second degree curve S passing through A, B, C, D can we written as 1 3 2 4 0 S L L L L ≡ + λ · Observe carefully that the substitution of the co-ordinates of any of the four points A, B, C, D will make both sides identically 0, implying that these four points lie on S. We now simply impose the necessary constraint ( ) onλto make S represent a circle, thus obtaining S ! Consider a circle of radius r centred at the origin: 2 2 2 x y r + · A line y = mx + c either just touches this circle or intersects it in two distinct points. What condition must m and c satisfy? Solution: What we need to do algebraically is solve the simultaneous system of equations 2 2 2 x y r + · ... (1) y mx c · + ... (2) and find the condition on m and c for this system to have two distinct roots. We substitute the value of y from (2) in (1): ( ) 2 2 2 x mx c r + + · ( ) 2 2 2 2 1 2 0 m x mc x c r ⇒ + + + − · ... (3) Since the line intersects the circle (or touches it) the discriminant of (3) cannot be non-negative since at least one real value of x must exist. Thus: ( )( ) 2 2 2 2 2 4 4 1 m c m c r ≥ + − 2 2 2 2 0 c r m r ⇒ − − ≤ ( ) 2 2 2 1 c r m ⇒ ≤ + This is the condition that m and c must satisfy. Incidentally, we also obtained the condition for tangency: ( ) 2 2 2 1 c r m · + 2 1 c r m ⇒ · t + Example – 08 L OCUS L OCUS L OCUS L OCUS L OCUS 10 Mathematics / Circles Thus, 2 1 y mx r m · t + will always be tangents to the circle x 2 + y 2 = r 2 , whatever be the value of m. A particular case that you should observe here is that when, m → ∞the equation becomes 2 2 lim 1 1 m y m x r m m →∞ ¹ ¹ · t ' ; + + ¹ ¹ x r ⇒ · t which means that the two tangents are vertical and touch the circle at the two end-points of the horizontal diameter. This is intuitively obvious. A point P moves in the Euclidean plane in such a way that, PA PB · λwhere A and B are fixed points and0 λ > . Find the locus of P. Solution: The easiest case is when 1; λ ·then PA = PB and P will hence lie on the perpendicular bisector of AB. We consider the case when1. λ ≠Let A and B be assigned the co-ordinates (a, 0) and (– a, 0) (for convenience). This can always be done by an appropriate choice of the co-ordinate axes. Now, let P have the co-ordinates (x, y). We have, 2 2 2 PA PB · λ ( ) ( ) { } 2 2 2 2 2 x a y x a y ⇒ − + · λ + + ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 2 1 1 0 x y ax x a ⇒ − λ + −λ − + λ + −λ · ( ) ( ) 2 2 2 2 2 1 2 0 1 x y a x a + λ ⇒ + − + · −λ ... (1) This is obviously the equation of a circle centred at ( ) ( ) 2 2 1 , 0 1 a ¸ _ +λ −λ ¸ , . Note that this circle does not pass through either A or B. Let us consider an example of this. Let AB be 2 units, so that we can assign (1, 0) and (–1, 0) as the co-ordinates A and B. Let P move in such a way that PA = 2PB,i.e,2. λ ·From (1), the locus of P is the circle: ( ) 2 2 2 1 4 1 0 1 4 x y x + + − + · − 2 2 10 1 0 3 x y x ⇒ + + + · Example – 09 L OCUS L OCUS L OCUS L OCUS L OCUS 11 Mathematics / Circles The centre of this circle is 5 , 0 3 ¸ _ − ¸ , and its radius is( ) 2 2 5 4 0 1 3 3 ¸ _ − + − · ¸ , Fig - 08 y A P (1, 0) B (–1, 0) 5 ,0 3 − For any point taken on the circumference of this circle, we will have 2 P PA =PB The circle that we obtained A fixed line L 1 intersects the co-ordinate axes at P(a, 0) and Q (0, b). A variable line L 2 , perpendicular to L 1 , intersects the axes at R and S. Show that the locus of the points of intersection of PS and QR is a circle. Solution: The equation of L 1 , using intercept form, can be written as 1 x y a b + · bx ay ab ⇒ + · Since L 2 is perpendicular to L 1 , its equation can be written as 2 0 L ax by ≡ − + λ · where λ is a real parameter. Using the equation of L 2 , we can determine R and S to be , 0 a λ ¸ _ − ¸ , and 0, b λ ¸ _ ¸ , respectively. Fig - 09 y x P a,( 0) Q , b (0 ) R a λ − S b λ 0, L 2 Points and are fixed whereas and will vary as varies P Q R S λ ,0 Example – 10 L OCUS L OCUS L OCUS L OCUS L OCUS 12 Mathematics / Circles We now write the equations to PS and QR using the two-point form: : y PS x aby a x a ab −λ · ⇒ λ + · λ − .... (1) : y b ab QR abx y b x − · ⇒ − + λ · λ λ .... (2) The relation that the intersection point of PS and QR, will satisfy can be evaluated by eliminating λ from (1) and (2). We thus obtain aby abx a x y b λ · · − − 2 2 2 2 aby aby a bx abx ⇒ − · − 2 2 0 x y ax by ⇒ + − − · This represents a circle centred at , 2 2 a b ¸ _ ¸ , and passing through the origin. Let 1 , , 1, 2, 3, 4 i i m i m ¸ _ · ¸ , be four distinct points lying on a circle. Prove that 1 2 3 4 1 m m m m· Solution: We first assume an equation for this circle C, in its general form: 2 2 : 2 2 0 C x y gx fy c + + + + · Since 1 , i i m m ¸ _ ¸ , satisfies the equation of C for i = 1, 2, 3, 4 we have 2 2 1 2 2 0 1, 2, 3, 4 i i i i f m gm c i m m + + + + · · 4 3 2 2 2 1 0 1, 2, 3, 4 i i i i m gm cm fm i ⇒ + + + + · · This last equation tells us that ' i m sare the roots of the following equation in m: 4 3 2 2 2 1 0 m gm cm fm + + + + ·: Roots of this equation are m, i = 1, 2, 3, 4 The product of the roots, which is 1 2 3 4 m m m m , can easily be seen to be 1 from this equation. Example – 11 L OCUS L OCUS L OCUS L OCUS L OCUS 13 Mathematics / Circles Find the equation of the circle C which has two fixed points( ) ( ) 1 1 2 2 , and Ax y Bxyas the end-points of its diameter. Solution: To evaluate the required equation, we can use a well known result from plane geometry: the angle in a semicircle is a right angle. Fig - 10 B x y ( ,) 2 2 For any point ( ,) on the circumference of the circle, the angle is a right angle Pxy APB P x y ( ,) A x y ( ,) 1 1 Thus, we can use this fact: (Slope of AP) × (Slope of PB) = – 1 1 2 1 2 1 y y y y x x x x − − ⇒ × · − − − ( )( ) ( )( ) 1 2 1 2 0 x x x x y y y y ⇒ − − + − − · ... (1) This is the required equation that will represent C. Note that we could equivalently have used the Pythagoras theorem inAPB ∆to evaluate the equation of C: 2 2 2 AP PB AB + · ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 2 2 1 2 1 2 x x y y x x y y x x y y ⇒ − + − + − + − · − + − ( ) ( ) 2 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 x y x x x y y y x x y y ⇒ + − + − + · − − ( )( ) ( )( ) 1 2 1 2 0 x x x x y y y y ⇒ − − + − − · which is the same as what we obtained in (1). Given the circle 2 2 : 2 2 0 C x y gx fy c + + + + · , find the intercepts that it makes on the x-axis and y-axis. Solution: We did a similar case in Example - 8 by solving simultaneously the equation of the circle and the line on which the intercept is required. Here, we’ll proceed analogously. Example – 12 Example – 13 L OCUS L OCUS L OCUS L OCUS L OCUS 14 Mathematics / Circles x – intercept Fig - 11 To find the-intercept, we put = 0 in the equation for : + 2 + = 0 ... (1) This gives two values for (real and distinct, equal or imaginary) and as shown alongside. We need to find |–| x y C x gxc x x x x x 2 1 2 1 2 x 1 x 2 x-axis C = 0 x-intercept From (1), we have 1 2 1 2 2 and . x x g x x c + · − · Thus, ( ) 2 1 2 1 2 1 2 4 x x x x x x − · + − 2 2 g c · − y - intercept Fig - 12 We put = 0 to get + 2 + = 0 ... (2) If this has roots,, the length of the intercept is x y fyc y y 2 1 2 y 1 y - axis y 2 y - intercept ( ) 2 1 2 1 2 1 2 4 y y y y y y − = + − ( ) ( ) 2 2 using 2 f c = - Thus, the intercepts are of length 2 2 2 and 2 g c f c − − respectively. Obviously, if 2 g c ... (2) 2 2 6 0 a ⇒ − > 3 or 3 a a ⇒ >< − ... (ii) The intersection of (i) and (ii) gives us the required values ofa as ( ) ( ) 3 2, 3 3, 3 2 a∈− − ∪ Findthelocusofthefootoftheperpendiculardrawnfromtheoriginuponanychordofacircle 2 2 2 2 0 S x y gx fy c ≡ + + + + ·which subtends a right angle at the origin. Solution: The situation is depicted graphically in the figure below to make things clearer: Fig - 13 O AB S X l m is the origin. is a particular chord of the circle which subtends a right angle at the origin. We need to find the locus of the foot of the perpendicular, i.e.( ,). B O y x X l m ( , ) A Example – 14 Example – 15 L OCUS L OCUS L OCUS L OCUS L OCUS 16 Mathematics / Circles Observe that the equation of the chord AB can be written as ( ) y m l AB OX x l m − − · ⊥ − ∵ 2 2 lx my l m ⇒ + · + Now, if we homogenize the equation of S using the equation of the chord AB, what we’ll get is the equation of the pair of straight lines OA and OB (as discussed in the last chapter on straight lines). This is what we proceed to do: 2 2 2 2 2 2 2 2 l 2 2 0 x my lx my lx my x y gx fy c l m l m l m 2 + + + ¸ _ ¸ _ ¸ _ + + + + · + + + ¸ , ¸ , ¸ , ... (1) This is the joint equation of OA and OB, since OA and OB need to be at right angles, we impose the appropriate constraint for perpendicularity on (1): 2 2 Coeff. ofcoeff .of 0 x y + · ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 0 gl cl fm cm l m l m l m l m ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ + + + + + · ' ; ' ; + + + + ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 0 cl m gl fm l m l m l m + ⇒ + + + · + + + 2 2 0 2 c l m gl fm ⇒ + + + + · This is the equation of a circle. To be more conventional, we should use (x, y) instead of the variables l and m. Thus, the required locus is 2 2 0 2 c x y gx fy + + + + · L OCUS L OCUS L OCUS L OCUS L OCUS 17 Mathematics / Circles Q. 1 Find the greatest and least distances of the point P (10, 7) from the circle 2 2 4 2 20 0 x y x y + − − − · Q. 2 Through the origin O, a straight line is drawn to cut the liney mx c · +at P. IfQ is a point on this line such that 2 OPOQ ⋅ · λ, show that the locus of Q is a circle passing through the origin. Q. 3 What is the area of an equilateral triangle inscribed in the circle 2 2 2 2 0? x y gx fy c + + + + · Q. 4 Determine the equation of the circle passing through the points (1, 2) and (3, 4) and touching the line 3 3 0. x y + − · Q. 5 A circle whose centre is the point of intersection of the lines2 3 4 0 x y − + ·and 3 4 5 0 x y + − · passes through the origin. Find its equation. Q. 6 Find the equation of the circle which touches the x-axis and two of whose diameters lie along 2 5 0 and 3 2 8 0 x y x y − − · − − · Q. 7 Find the equation of the circumcircle of an equilateral triangle two of whose vertices are (–1, 0) and (1, 0) and the third vertex lies above the x–axis. Q. 8 Find the equation of the circle passing through (1,0) and (0,1) and having the smallest possible radius. Q. 9 The equations of the sides of a quadrilateral are given by( ) 0 1, 2, 3, 4 . r r r r L a x b y c r · + + · · If the quadrilateral is concyclic, show that 1 3 1 3 1 3 3 1 2 4 2 4 2 4 4 2 a a b b a b a b a a b b a b a b − + · − + Q. 10 Find the point on the straight line2 11 y x · +which is nearest to the circle ( ) 2 2 16 32 8 50 0 x y x y + + − − · TRY YOURSELF - I L OCUS L OCUS L OCUS L OCUS L OCUS 18 Mathematics / Circles We first consider the case which we’ve already considered in Example - 8 earlier, namely, when will the line y mx c · + be a tangent to the circle 2 2 2 ? x y a + · Earlier, we solved the simultaneous system of two equations and put the discriminant of the resulting quadratic equal to 0 to obtain the condition for tangency. Here, we follow an alternative approach. We use the simple geometric fact that if a line L is a tangent to a circle S, then the perpendicular distance of the centre of S from L must equal the radius of S: Fig - 14 L = 0 C If is tangent to , then must equal, the radius of. L S CP r S S = 0 P In the current case, the centre of 2 2 2 x y a + ·is (0, 0) and its radius is a. The perpendicular distance of (0, 0) fromy mx c · +must equal a, i.e. 2 1 c a m · + ⇒ 2 2 2 (1 ) c a m · + This is the same condition that we derived earlier. From this relation, we can infer that any line of the form 2 1 y mx a m · t + will always be a tangent to the circle 2 2 2 , x y a + ·whatever the value of m may be. Consider now the problem of finding the equation of the tangent to the circle 2 2 2 x y a + ·at a point 1 1 ( , ) Px y lying on the circle. Fig - 15 We want to find the equation of the tangent at ( , ) lying on the circle + = T Pxy x y a 1 1 22 2 P x y ( , ) 1 1 O, (0 0) T x x Sec t i on-2 TANGENTS AND CHORDS L OCUS L OCUS L OCUS L OCUS L OCUS 19 Mathematics / Circles The tangent T will obviously be perpendicular to OP. Thus, the equation of T is 1 1 1 1 y y x x x y − − · − 2 2 1 1 1 1 xx yy x y ⇒ + · + ...(1) The other piece of information that we have is that 1 1 ( , ) x ylies on the circle and therefore must satisfy its equation. Thus, 2 2 2 1 1 . x y a + ·Using this in (1), we obtain the equation of T as 2 1 1 : T xx yy a + · If the point 1 1 ( , ) x yhas been specified in polar form, i.e. in the form ( cos , sin ), a a θ θthe equation of T becomes : cos sin T x y a θ θ + · Wenowgotothegeneralcaseoffindingtheequationofthetangenttoanarbitrarycircle 2 2 2 2 0 S x y gx fy c ≡ + + + + ·at a point 1 1 ( ) Px ylying on this circle: Fig - 16 We want to find the equation of the tangent at ( ,) lying on the circle. T Px y S 1 1 Px y (,) 1 1 C g,- f(- ) T x y Here again, as in the earlier case, we have two pieces of information which we can put to use : P lies on S : 2 2 1 1 1 1 2 2 0 x y gx fy c + + + + · ...(1) T is perpendicular to CP : 1 1 1 1 1 y y y f x x x g − + × · − − + { } ( , ) is any point onxy T 2 2 1 1 1 1 1 1 xx yy gx fy x y gx fy ⇒ + + + · + + + We now add 1 1 ( ) gx fy c + +on both sides above : L OCUS L OCUS L OCUS L OCUS L OCUS 20 Mathematics / Circles 2 2 1 1 1 1 1 1 1 1 ( ) ( ) 2 2 xx yy gx x f y y c x y gx fy c ⇒ + + + + + + · + + + + Now we use (1) for the RHSabove to finally obtain the equation of T as 1 1 1 1 : ( ) ( ) 0 T xx yy gx x f y y c + + + + + + · The expression on the left hand side of the equation above is conventionally denoted as 1 1 ( , ) Tx y (Note that T is a function of x and y too). Thus, the equation of the tangent can be written concisely as 1 1 1 1 ( , ) 0 : Equation of tangent at ( , ) toTx y x y S · Usingananalogousapproach,wecanwritetheequationofthenormaltothecircle 2 2 2 2 0 x y gx fy c + + + + ·at the point( ) 1 1 , Px y . You are urged to do this yourself. Note that every normal of a circle willpass through the circle’s centre. Find the equation of the tangent to (a) 2 2 1 x y + ·at 1 2 , 3 3 ¸ _ ¸ , (b) 2 2 2 4 4 0 x y x y + − − + ·at 1 3 , 2 2 2 ¸ _ + ¸ , Solution (a) The equation of the tangent to 2 2 2 x y a + ·at 1 1 ( , ) x yis, as obtained in the preceeding discussion, 1 1 ( , ) 0 Tx y · 2 1 1 xx yy a ⇒ + · Here, 1 1 1 2 ( , ) , 3 3 x y ¸ _ · ¸ , and1 a · . Thus, the required equation is 2 3 x y + · (b) The equation of the tangent to 2 2 2 2 0 x y gx fy c + + + + ·at 1 1 ( , ) x yis 1 1 ( , ) 0 Tx y · 1 1 1 1 ( ) ( ) 0 xx yy gx x f y y c ⇒ + + + + + + · Here, 1 1 ( , ) x yis 1 3 , 2 2 2 ¸ _ + ¸ , and thus, the required equation is 3 1 3 2 1 2 2 4 0 2 2 2 2 x y x y ¸ _ ¸ _ ¸ _ + + − + − + + + · ¸ , ¸ , ¸ , 3 1 3 0 2 2 2 x y − ¸ _ ⇒ + − + · ¸ , 3 (2 3 1) 0 x y ⇒ − + − + · Example – 16 L OCUS L OCUS L OCUS L OCUS L OCUS 21 Mathematics / Circles What is the length of the tangent to 2 2 2 2 0 S x y gx fy c ≡ + + + + ·drawn from an external point 1 1 ( , ) Px y ? Solution : Fig - 17 O g fS Px y PA PB P S PA PB (- , - ) is the centre of. ( ,) is an external point. Note that two tangents, and can drawn from to. Also, = 1 1 A O g,-f(- ) B P x ,y ( ) 1 1 The length PA (or PB) can be evaluated by a simple application of the Pythagoras theorem. In, PAO ∆observe that 2 2 2 PA PO AO · − { } { } 2 2 2 2 2 1 1 ( ) ( ) x g y f g f c · + + + − + − { } is the radius AO ∵ 2 2 1 1 1 1 2 2 x y gx fy c · + + + + ...(1) The equation of the circle being represented by0, S ·we can denote the RHS obtained in (1) by 1 1 ( , ). Sx yThus, the length of the tangent can be written concisely as 1 1 ( , ) PA Sx y · For example, the length of the tangent from (4, 4) to 2 2 2 4 4 0 x y x y + − − + ·will be 2 2 4 4 2 4 4 4 4 l · + − × − × + 12 · 2 3 · In the next example, we discuss how to write the equation to the pair of tangents PA and PB. From an external point 1 1 ( , ), Px y(two) tangents are drawn to the circle 2 2 2 2 0 S x y gx fy c ≡ + + + + · . These tangents touch the circle at A and B. Find the joint equation of PA and PB. Example – 17 Example – 18 L OCUS L OCUS L OCUS L OCUS L OCUS 22 Mathematics / Circles Solution : Consider any point ( , ) h klying on the tangents drawn from P to S. Fig - 18 Assume ( ) to lie on one of the two tangents or h, k PA PB A (- ) g,-f B P x ,y ( ) 1 1 ( ) h k 1 Since we know two points on the line PA, we can use the two-point form to write its equation : 1 1 y k y k x h x h − − · − − 1 1 1 1 ( ) ( ) ( ) ( ) xy k yx h hy k kx h ⇒ − − − · − − − 1 1 1 1 ( ) ( ) ( ) 0 xy k yx h kx hy ⇒ − − − + − · Since PA is a tangent to S, its distance from the centre of the circle, ( , ), g f − −must equal the radius. This gives 2 2 2 1 1 1 1 2 2 1 1 ( ( ) ( ) ) ( ) ( ) gy k f x h kx hy g f c x h y k − − + − + − · + − − + − ...(1) To write the equation of the pair of lines in conventional form, we use ( , ) xyinstead of ( , ) h kin (1), above. Subsequent (lengthy !) rearrangements give: { } ( )( ) 2 2 2 2 2 1 1 1 1 1 1 1 1 ( ) ( ) 2 2 2 2 xx yy gx x f y y c x y gx fy c x y gx fy c + + + + + + · + + + + + + + + The left hand side can be written concisely as 2 1 1 ( ( , )) Tx yas described earlier whereas the right hand side can be written concisely as 1 1 ( , ) ( , ). Sxy Sx yThus, the equation to the pair of tangents can be written concisely as 2 1 1 1 1 ( , ) ( , ) ( , ) T x y SxySx y · This relation be written in an even shorter form as simply 2 1 T SS · . L OCUS L OCUS L OCUS L OCUS L OCUS 23 Mathematics / Circles Find the equation to the pair of tangents drawn from the origin to the circle 2 2 4 4 7 0 x y x y + − − + · . Solution: We use the relation obtained in the last example, 2 1 , T SS ·to write the desired equation. Here, 1 1 ( , ) x yis (0, 0) while2, 2 g f · − · −and7 c · . Thus the joint equation is 2 (0, 0) ( , ) (0, 0) T SxyS · 2 2 2 ( 2 2 7) ( 4 4 7)(7) x y x y x y ⇒ − − + · + − − + 2 2 2 2 4 4 49 8 28 28 7 7 28 28 49 x y xy x y x y x y ⇒ + + + − − · + − − + 2 2 3 8 3 0 x xy y ⇒ − + · As expected, since the tangents have been drawn from the origin, the obtained equation is a homogenous one. Refer to Fig - 18. Suppose that A and B, the points of contact of the two tangents, are joined. What will be the equation to this chord of contact ? Solution: Assume the co-ordinates of A and B to be ( , ) A A x yand ( , ) B B x yrespectively. The equations of the tangents at A and B are, respectively, } ( ) ( ) 0 ( ) ( ) 0 A A A A B B B B xx yy gx x f y y c xx yy gx x f y y c + + + + + + · + + + + + + · ...(I) Since the two tangents intersect at P, the co-ordinates of P must satisfy the system of equations I. Thus, } 1 1 1 1 1 1 1 1 ( ) ( ) 0 ( ) ( ) 0 A A A A B B B B x x y y gx x f y y c x x y y gx x f y y c + + + + + + · + + + + + + · ...(II) Now, if you observe the system II carefully, you will realise that we can think of it this way : ( , ) A A x y and ( , ) B B x yare two points which satisfy the linear equation 1 1 1 1 ( ) ( ) 0 xx yy gx x f y y c + + + + + + · ...(1) This is because substitution of( , ) A A x yor( , ) B B x yinto (1) results in the system of equations II. Thus (1) must be the equation of the chord of contact we are looking for, since both A and B satisfy (1). Example – 19 Example – 20 L OCUS L OCUS L OCUS L OCUS L OCUS 24 Mathematics / Circles We can write this obtained equation concisely as 1 1 ( , ) 0 Tx y ·: Equation of the chord of contact As an example, the chord of contact for the two tangents drawn from theorigin to the circle 2 2 4 4 7 0 x y x y + − − + ·will be 1 1 1 1 ( ) ( ) 0 xx yy gx x f y y c + + + + + + ·where 1 1 ( , ) (0, 0) x y · 2 2 7 0 x y ⇒ − − + · 7 2 x y ⇒ + · From an external point P, a line is drawn intersecting a circle S in two distinct points A and B. A tangent is also drawn from P touching the circle S at T. Prove that PAPB ⋅ is always constant, and equal to 2 . PT Solution: Let the equation of the circle be 2 2 2 2 0 S x y gx fy c ≡ + + + + ·and the point P be 1 1 ( , ) x y : Fig - 19 We need to prove that= PAPB PT • 2 P x ,y ( ) 1 1 B A S T Notice that the line PABpasses through the fixed point P. What is variable about it is its slope, which we assume to be tanθ . Thus, using the polar form for the lines, we obtain the equation of PABas 1 1 cos sin x x y y r θ θ − − · · ...(1) In particular, the co-ordinates of A and B can be obtained in terms of θ using the value of r as PA and PB respectively in (1). Using (1), we can write any point on the line PAB as 1 1 ( cos , sin ). x r y r θ θ + +If this point lies on the circle, it must satisfy the circle’s equation : 2 2 1 1 1 1 ( cos ) ( sin ) 2 ( cos ) 2 ( sin ) 0 x r y r gx r f y r c θ θ θ θ + + + + + + + + · Example – 21 L OCUS L OCUS L OCUS L OCUS L OCUS 25 Mathematics / Circles 2 2 2 1 1 1 1 (2 cos 2 sin ) 2 2 0 r g f r x y gx fy c ⇒ + + + + + + + · θ θ ...(2) The equation (2) in r will have two roots 1 rand 2 r corresponding to PA and PB since A and B lie on the circle. Thus, 2 2 1 2 1 1 1 1 2 2 PAPB r r x y gx fy c ⋅ · ⋅ · + + + + 1 1 ( , ) Sx y · 2 PT · (using the result obtained in Example - 17) You are urged to prove this result using ‘pure’ geometry. From an external fixed point( , ), P h ktangents are drawn to the circle 2 2 2 x y a + · . Find the area of the triangle formed by these tangents and their chord of contact. Solution: Fig - 20 AB P x y a PAB AB PC is the chord of contact for the tangents drawn from to + = Observe that the area of can be written as • 2 2 2 ∆ • P h,k ( ) B A C (0 0) , O 1 2 Observe that OAC ∆ is similar to OAP ∆ . Thus, the ratio of the corresponding sides is equal. OC AC OA OA AP OP · · OA is simply the radius a, while OP is 2 2 . h k + Thus, 2 2 2 2 OA a OC OP h k · · + Example – 22 L OCUS L OCUS L OCUS L OCUS L OCUS 26 Mathematics / Circles Now, AC can be evaluated using Pythagoras theorem in : OAC ∆ 2 2 AC OA OC · − 4 2 2 2 a a h k · − + 2 2 2 2 2 h k a a h k + − · + The area ofPAB ∆is 1 2 AB PC ∆ · × × 1 (2 ) ( ) 2 AC OP OC · × × − 2 2 2 3/ 2 2 2 ( ) a h k a h k + − · + ____________________________________________________________________________________ Observe that this question was solved mostly through ‘pure’ geometrical considerations. A pure co-ordinate approach for the area of the triangle can be followed but you can expect it to be much longer. Thus, in this subject, you have to put your intuitive skills to the best use possible to determine the shortest approach. Let us consider another example here itself which will show why a pure geometrical approach is better sometimes than using co-ordinates. Consider two circles with the following equations : 2 2 1 2 2 0 S x y gx fy c ≡ + + + + · 2 2 2 2 2 2 2 2 2 sin ( ) cos 0 S x y gx fy c g f α α ≡ + + + + + + · These two circles are obviously concentric, centred at ( , ). g f − −The radius of 1 Sis 2 2 1 r g f c · + − while that of 2 Sis 2 2 2 1 sin sin r g f c r α α · + − · so that 1 2 . r r > The problem is as follows : from any point on 1 , Stwo tangents are drawn to 2 . S What is the angle between these two tangents. L OCUS L OCUS L OCUS L OCUS L OCUS 27 Mathematics / Circles Obviously, this angle should be the same for any point chosen on 1 Ssince the circles are perfectly symmetrical figures. Let us draw a diagram corresponding to this situation : Fig - 21 Note that = = = sin Thus, = PO r OT r r OPT 1 2 1 α ⇒ ∠ ∠ α α TPS =OPT 2= 2 O T P S As explained in the diagram, the angle between the two tangents can be evaluated by simple geometric considerations to be 2 . α Although this example is more or less trivial, the contrast between pure geometric and co-ordinate approach will become more apparent in some subsequent examples. A point P moves in such a way so that the tangents drawn from it to the circle 2 2 2 x y a + ·are perpendicular. Find the locus of P. Solution: To contrast between the various alternatives available to us, we will use all of them here: A PURE-GEOMETRIC APPROACH: Fig - 22 T P S O (0,0) a From the figure, it is apparent that OTPS is a square since all the angles are right angles and OS = OT = a . Thus, 2 OP a · , i.e., the distance of P from O is always 2 , a i.e., P lies on a circle of radius. 2 a .Thus, P satisfies the equation 2 2 2 2 x y a + · This circle is called the Director circle of the given circle. A CO-ORDINATE APPROACH - I Let P be the point ( , ). h kThe equation of the pair of tangents drawn from P to the circle is 2 1 T SS · 2 2 2 2 2 2 2 2 ( ) ( )( ) hx ky a x y a h k a ⇒ + − · + − + − Example – 23 DI RECTOR CI RCLE L OCUS L OCUS L OCUS L OCUS L OCUS 28 Mathematics / Circles This combined equation will represent a pair of perpendicular straight lines if 2 2 Coeff. ofCoeff. of0 x y + · 2 2 2 2 ( ) ( ) 0 k a h a ⇒ − + − · 2 2 2 2 h k a ⇒ + · Using ( , ) xyinstead of ( , ), h kwe obtain the locus of P in conventional form : 2 2 2 2 x y a + · A CO-ORDINATE APPROACH - II The equation of any tangent to the circle 2 2 2 x y a + ·can be written as 2 1 . y mx a m · + + If this line passes through( , ), P h kthe co-ordinates of P must satisfy this equation: 2 1 k mh a m · + + 2 2 2 ( ) (1 ) k mh a m ⇒ − · + 2 2 2 2 2 ( ) 2 ( ) 0 h a m mhk k a ⇒ − − + − · This is a quadratic in m and will yield two values, say 1 mand 2 , mwhich physically corresponds to the fact that two tangents can be drawn from( , ) P h kto the circle. The two tangents are at a right angle if 1 2 1 mm· − 2 2 2 2 1 k a h a − ⇒ · − − 2 2 2 2 h k a ⇒ + · Using ( , ) xyinstead of( , ) kh kwe obtain the locus of P: 2 2 2 2 x y a + · If this question were to be encountered in an exam, the pure-geometric approach would certainly turn out to be the fastest ! 1 Cand 2 Care two concentric circles, the radius of 2 Cbeing twice that of 1 C . From a point P on 2 , Ctangents PA andPB are drawn to 1 . CProve that the centroid ofPAB ∆is on 1 . C Example – 24 L OCUS L OCUS L OCUS L OCUS L OCUS 29 Mathematics / Circles Solution: Let us again attempt this question using both a pure-geometric and a co-ordinate approach. PURE-GEOMETRIC APPROACH: Recall the following straightforward theorem pertaining to right-angle triangles. D C B A Fig - 23 ∆ ABC is right-angled at B and BD is the median drawn from B to the opposite side AC. Then our theorem tells us that BD = AD = CD This can be proved using simple geometry We will put this theorem to use in the current example. P A B D O C 1 Fig - 24 C 2 E Since the radius r 2 of C 2 is twice that of C 1 (r 1 ) we have PO = 2 OD ⇒ OD = PD. ... (1) Thus,D is the mid-point of PO. This means that in ∆ OAP, AD is the median to OP. By the theorem mentioned above, we have AD = OD = PD = r 1 Thus, in quadrilateral, ADBOwe have 1 . AD BD OA OB r · · · ·In other words,ADBO is a parallelogram soEis the midpoint of. ODThus, 1 1 / 2 2 ED r OD · ·...(2) Also, since, AE EB · PE is the median of . PAB ∆Thus the centroid lies on. PE From (1) and (2), we finally obtain2 . PD ED ·Thus, D divides the median PE in the ratio 2 : 1 implying D is the centroid which lies on 1 . C L OCUS L OCUS L OCUS L OCUS L OCUS 30 Mathematics / Circles The descriptive nature of the pure-geometric solution just provided might make it appear to be very long but actually only a few simple elementary geometry facts have been used. CO-ORDINATE APPROACH There’s no loss of generality in assuming that the two circles are centred at the origin. Thus, we can write their equations as 2 2 2 1 2 2 2 2 : : 4 C x y r C x y r + · + · Assume the point P to have the co-ordinates ( , ). h kThe equation of AB (the chord of contact) can then be writtenas Equation of 2 : AB hx ky a + · We can evaluate the co-ordinates of 1 1 ( , ) A x yand 2 2 ( , ) Bx yby simultaneously solving the equations for 1 Cand AB. Thus, 1 xand 2 xwill be the roots of 2 2 2 2 r hx x r k ¸ _ − + · ¸ , 2 2 2 2 2 2 2 ( ) 2 ( ) 0 h k x rhx r r k ⇒ + − + − · 2 1 2 2 2 2rh x x h k ⇒ + · + and 2 2 2 1 2 2 2 ( ) r r k x x h k − · + ... (3) 1 yand 2 ywill be the roots of 2 2 2 2 r ky y r h ¸ _ − + · ¸ , 2 2 2 2 2 2 2 ( ) 2 ( ) 0 h k y rky r r h ⇒ + − + − · 2 1 2 2 2 2rk y y h k ⇒ + · + and 2 2 2 1 2 2 2 ( ) r r h y y h k − · + ... (4) If we let ( , ) t sbe the co-ordinates of the centroid G of, PAB ∆we have ( ) ( ) 2 1 2 2 2 2 1 2 2 2 2 3 1 3 Using 3 and 4 2 3 1 3 x x h r t t h h k y y k r s s k h k ¹ ¸ _ + + · ⇒ · + ¹ + ¸ , ¹ ¹ ; ¹ ¸ _ + + · ⇒ · + ¹ + ¹ ¸ , ¹ Finally, observe that 2 2 2 2 2 2 2 2 2 9( ) ( ) 1 r t s h k h k ¸ _ + · + + + ¸ , But since ( , ) h klies on 2 , Cwe have 2 2 2 4 . h k r + · L OCUS L OCUS L OCUS L OCUS L OCUS 31 Mathematics / Circles Thus, 2 2 2 2 9 9( ) (4 ) 9 4 t s r r ¸ _ + · · ¸ , 2 2 2 t s r ⇒ + · implying that( , ) G t slies on 1 . C Since now you are in a good position to compare the two approaches, which one could you have rather chosen for this question, the pure-geometric one or the co-ordinate one! Consider the circle 2 2 2 . x y a + ·A chord of this circle is bisected at the point 1 1 ( , ). Px yWhat is the equation of this chord ? Solution : Convince yourself that such a chord will be unique, since it must be perpendicular to the line joining the origin to 1 1 ( , ), Px yas is clear from the figure below: O (0,0) A B P xy ( ) 11 Fig - 25 The chord is bisected at AB P x y OP AB. ( , ). Thus, we must have 1 1 ⊥ The slope of AB then becomes 1 1 x y − so its equation can be written simply as 1 1 1 1 ( ) x y y x x y − · − − 2 2 1 1 1 1 xx yy x y ⇒ + · + To make this equation look “better”, we subtract 2 a from both sides 2 2 2 2 1 1 1 1 xx yy a x y a + − · + − so that it can now be written concisely as 1 1 1 1 ( , ) ( , ) Tx y Sx y · Of course, this is easily generalised to the case when the equation of the circle is in the general form 2 2 2 2 0; x y gx fy c + + + + ·the result obtained is the same. The next example discusses a good application of this concept. Example – 25 L OCUS L OCUS L OCUS L OCUS L OCUS 32 Mathematics / Circles Find the locus of the mid-point of the chords of the circle 2 2 2 x y a + ·which subtend a right angle at the centre. Solution: Fig - 26 A O, (0 0) y M ( ) h ,k B x AB Mhk AB M is a chord of the circle which subtends a right angle at the centre. ( ,) is the mid-point of. We need to find the locus of. Since AB is bisected at( , ), Mh kwe can use the result obtained in the last example to write the equation of AB: ( , ) ( , ) Th k Sh k · 2 2 2 2 hx ky a h k a ⇒ + − · + − 2 2 hx ky h k ⇒ + · + We can view the chord AB as a line intersecting the curve (circle) 2 2 2 . x y a + ·Thus, we can obtain the joint equation of OA and OB by homogenizing the equation of the circle using the equation of the chord AB: 2 2 2 2 2 2 0 hx ky x y a h k + ¸ _ + − · + ¸ , Joint equation of OA and OB: 2 2 2 2 2 2 2 ( ) ( ) ( ) 0 h k x y a hx ky ⇒ + + − + · ...(1) Since OA and OB are perpendicular, we must have 2 2 Coeff .ofCoeff. of0 x y + · in (1) 2 2 2 2 2 2 2 2( ) 0 h k a h a k ⇒ + − − · 2 2 2 2 a h k ⇒ + · Using ( , ) xyinstead of ( , ), h kwe obtain the following equation as the locus of M: 2 2 2 2 a x y + · This is a circle concentric with the original circle as might have been expected. Try solving this question using pure geometric considerations; the solution will be much simpler. Example – 26 L OCUS L OCUS L OCUS L OCUS L OCUS 33 Mathematics / Circles Consider two circles with the following equations: 2 2 1 : 2 2 1 0 C x y x y + − − + · 2 2 2 : 16 2 61 0 C x y x y + − − + · Find the values that a can take so that the variable line2 y x a · +lies between these two circles without touching or intersecting either of them. Solution : Observe carefully that what is variable about the variable line2 y x a · +is not its slope but its y-intercept a. Thus, we can always adjust a so that line stays between the two circles. The following diagram makes this clear. Fig - 27 y x The lines can vary in this range so that they stay between the two circles. Thus as long on < < the line = 2+ stays between the two circles a a a y x a min max (1,1) (8,1) C 1 (0,) a max C 2 (0,) a min Evaluate max a: The line max 2 y x a · +is a tangent to 1 Cif the perpendicular distance of the centre (1, 1) of 1 Cfrom this line is equal to ' 1 C sradius which is 1. Thus : max 2 1 1 5 a − + · max 1 5 a ⇒ + · t max 5 1 a ⇒ · − − max since from the figure we can see thatis definitely negative a ¹ ¹ ¹ ¹ ' ; ¹ ¹ ¹ ¹ Example – 27 L OCUS L OCUS L OCUS L OCUS L OCUS 34 Mathematics / Circles Evaluate min a: The distance of ' 2 Cscenter (8, 1) from min 2 y x a · +must be equal to its radius which is equal to 2. Thus : min 16 1 2 5 a − + · min 15 2 5 a ⇒ + · t min 2 5 15 a ⇒ · − min min 2 we have selected the larger of the two values possible since that is what corresponds to, i.e. because the line2lies. a y x a above C ¹ ¹ ¹ ¹ ' ; ¹ ¹ · + ¹ ¹ Thus, we obtain the allowed values of a as 2 5 15 5 1 a − < < − − A circle touches the liney x ·at a point P such that4 2 OP ·where O is the origin. The circle contains the point (–10, 2) in its interior and the length of its chord on the line0 x y + ·is 6 2. Determine the equation of the circle. Solution: As always, before starting with the solution, it is a good to draw a diagram of the situation described to get a feel of it. Also, as far as possible, we should try to use pure-geometric considerations to cut down on the (complicated) algebraic manipulations that would result otherwise. Fig - 28 y x It should be more or less apparent that to be able to contain the point (-10, 2) inside it, the centre of the circle must lie somewhere in the shaded region. The point is then (- 4, - 4) since = 42 Assume the centre to be at ( ,). The radius of this circle is then given by = where = (+ 4)+ (+ 4) P OP X hk r XP r h k 2 2 2 6 2 B A P(-4,-4) O x + y = 0 y = x X ' We have, ( ) PX y x ⊥ · 4 1 4 k h + ⇒ · − + ...(1) 8 0 h k ⇒ + + · ...(2) Example – 28 L OCUS L OCUS L OCUS L OCUS L OCUS 35 Mathematics / Circles Also, { } 2 2 2 2 Perpendicular dist. of from0 BX AX AB r AB X x y · · − · − + · 2 2 2 ( 4) ( 4) (3 2) 2 h k h k + ⇒ · + + + − ...(3) Using (1) and (2) in (3), we obtain 2 2 (4 2) 2( 4) 18 h · + − 4 5 h ⇒ + · t 9,1 h ⇒ · − Given the region in which X lies, h must be –9. Thus, from (2), k is 1 and the radius r is 5 2 . The required equation is therefore 2 2 2 ( 9) ( 1) (5 2) x y + + − · 2 2 18 2 32 0 x y x y ⇒ + + − + · ...(4) For students used to rigor, it can finally be verified that the circle given by (4) does indeed contain the point (-10, 2) and thus our initial assumption of the region in which the centre X lies, was correct. Let 1 Cand 2 Cbe two circles with 2 Clying inside 1 . CA circle C lying inside 1 Ctouches 1 Cinternally and 2 C externally. Determine the locus of the centre of C. Solution: First of all, you must note that 2 Cis not concentric with 1 . CAll that is said is that 2 Clies somewhere inside 1 . C Let us first discuss what all would be involved in solving this question through a co-ordinate approach. We could first assume a co-ordinate axes, say, with the x-axis lying along the line joining the centres of 1 Cand 2 Cand the origin at the centre of 1 : C C 1 O(0,0) C 2 x (a, 0) y The centre ofcan be assumed to be ( , 0) as shown. The equations of and can then be written as C a C C 2 1 2 C x+ y r 1 1 2 := 2 2 C x – a y r 2 2 : ( ) + = 22 2 Fig - 29 Example – 29 L OCUS L OCUS L OCUS L OCUS L OCUS 36 Mathematics / Circles We could then assume the equation of C to be 2 2 2 2 0 x y gx fy c + + + + ·and impose the necessary constraints on g, f and c so that C touches 2 Cexternally and 1 Cinternally. Recall the appropriate constraints for two circles 1 Sand 2 Stouching externally and internally. r 2 O O r r 1 2 1 2 = + Fig - 30 r 1 O 2 (Circles touch externally) O O r r 1 2 1 2 =| –| (Circles touch internally) O 1 r 1 O 2 r 2 O 1 Imposing these constraints gives us the necessary conditions that g, f and c must satisfy and hence the locus of thecentre of C which is ( , ). g f − − However, we will be much letter off in using a pure-geometry approach here also as you’ll soon see. Assume an arbitrary circle C of radius r inside 1 Cwith centre X, and which satisfies the given constraint: Fig - 31 C O 2 B O 1 A C 1 The circle touches internally at andexternally at. is the centre of while, and respectively. The radii of , and are, and respectively. C C A C BX C O C CC C rr r 1 2 1 2 1 2 1 2 O C 2 1 are centres of and C 2 X By the properties of circles touching internally and externally, we have 1 1 1 O X O A AX r r · − · − 2 2 2 O X O B BX r r · + · + 1 2 1 2 O X O X r r ⇒ + · + ...(1) (1) simply states the centre of C i.e. X, moves in such a way so that the sum of its distances from 1 O and 2 Ois constant. Thus, X must lie on an ellipse with 1 Oand 2 Oas the two foci ! To sketch the path that X can take, we can follow the approach described in the unit on Complex numbers. Fix two pegs at 1 Oand 2 Oand tie a string of length 1 2 r r +between these two pegs. Use your pencil and the taut string as a guide to trace out the ellipse. This is the path on which the centre of C can move. L OCUS L OCUS L OCUS L OCUS L OCUS 37 Mathematics / Circles Through an arbitrary fixed point 1 1 ( , ), Px ya variable line is drawn intersecting the circle 2 2 2 x y a + ·at A and B respectively. Tangents drawn to this circle at A and B intersect at Q. Find the locus of Q. Solution : Fig - 32 A B Q hk ( ) 1 The tangents at and to the circle + = intersect at . We wish to determine the locus of whose co-ordinates we have assumed to be ( ) A B x y a Q Q hk 2 2 2 1 Px ,y ( ) 1 1 We can view this situation from a different perspective. We have a point( , ) Qh kfrom which we draw tangents QA and QB to the circle 2 2 2 . x y a + ·Thus, the equation of AB (which is the chord of contact) is ( , ) 0 Th k · 2 hx ky a ⇒ + · Now, this chord of contact also passes through 1 1 ( , ) Px yso that 2 1 1 hx ky a + · ...(1) What we have in (1) is a linear equation involving the variables h and k. Note that 1 xand 1 yare constant. Thus, we can infer from (1) that ( , ) h klies on the line 2 1 1 xx yy a + · ...(2) This is the required equation ! The line obtained in (2) is referred to as the polar of the point P with respect to the given circle. P is itself referred to as the pole of the polar. Notice that the equation of the polar can be written concisely as 1 1 1 1 ( , ) 0 : Equation of the polar for the pole( , ) Tx y Px y · This example should show you that sometimes enormous simplifications are achieved using a co-ordinate geometrical approach rather than a pure-geometrical one. Co-ordinate geometry is not all that bad! Example – 30 POLE AND POLAR L OCUS L OCUS L OCUS L OCUS L OCUS 38 Mathematics / Circles Q. 1 Three circle with radii 1 2 , r rand 3 rtouch each other externally. The tangents at their points of contact meet at a point whose distance from any point of contact is 4. Show geometrically that 1 2 3 1 2 3 16 r r r r r r · + + Q. 2 What is the equation of the tangent to 2 2 30 6 109 0 x y x y + − + + ·at 1 (4 1) ? − Q. 3 Find the tangents to 2 2 6 4 12 0 x y x y + − + − ·parallel to4 3 5 0. x y + + · Q. 4 Find the tangents to 2 2 2 x y a + ·which make a triangle of area 2 a with the axes. Q. 5 If( , ) P a band( , ) Q b aare two points ( ), b a ≠find the equation of the circle touching OP and OQ at P and Q where O is the origin . Q. 6 Find the equation of the normal to 2 2 3 3 4 6 0 x y x y + − − ·at (0, 0) Q. 7 The line 2 1 0 x y − + ·is a tangent to a circle at (2, 5); its centre lies on9. x y + ·Find its equation. Q. 8 If 3 0 x y + ·is a tangent to a circle whose centre is (2, –1), find the other tangent to the circle from the origin. Q. 9 Findthelocusofthepointofintersectionoftangentstothecircle 2 2 2 2 4 6 9sin 13cos 0 x y x x + + − + α+ α ·which are inclined at an angle of 2α to each other. Q. 10 Prove that the intercept of the pair of tangents from the origin to 2 2 2 2 2 0, x y gx fy k + + + + ·on the liney · λ is2 2 2 k k g λ − . TRY YOURSELF - II L OCUS L OCUS L OCUS L OCUS L OCUS 39 Mathematics / Circles Consider two circles 1 Cand 2 Cwhich intersect in two distinct points A and B. Our purpose is to determine the equation of AB which will be termed the common chord of 1 Cand 2 . C Fig - 33 A B C 2 C 1 AB C C C S S is the common chord of the circles and. The equations of and C are given to be= 0 and = 0 respectively. 1 2 1 2 1 2 Let the equations of the circles be 2 2 1 1 1 1 2 2 0 S x y g x f y c ≡ + + + + · 2 2 2 2 2 2 2 2 0 S x y gx f y c ≡ + + + + · Consider the equation 1 2 0, S S S ≡ − ·i.e. 1 2 1 2 1 2 (2 2 ) (2 2 ) 0 S g g x f f y c c ≡ − + − + − · ...(1) Clearly, S is the equation of a straight line. Observe carefully that since A and B lie on both 1 Cand 2 C , the co- ordinates of A and B satisfy both 1 Sand 2 Sand thus 1 2 0. S S S ≡ − ·This means that 1 2 0 S S − ·represents a straight line passing through A and B. Thus, this is precisely the commonchord ! 1 2 0 : Equation of the common chord S S S ≡ − · You are urged to observe one important fact about S. It is perpendicular to the line joining the centres of 1 Cand 2 ; Cthis is but expected. Fig - 34 A C 2 C 1 Slope of O 1 O 2 is P B O 1 O 2 ( ) –g , 2 –f 2 ( ) –g , 1 –f 1 2 1 1 2 1 m f f g g − = − Slope of AB is from (1) 2 1 2 2 1 m f f g g − = − Thus, 1 m 2 m= – 1 The perpendicularity of 1 2 OOand AB can be proved geometrically too in a straightforward manner. Sec t i on-3 EQUATI ONS DESCRI BI NG CI RCLES L OCUS L OCUS L OCUS L OCUS L OCUS 40 Mathematics / Circles The length of the common chord can be easily evaluated using the Pythagoras theorem: 2 2 2 2 1 1 2 2 2 2 AB O A O P O A O P · − · − where 1 OP and 2 O Pare the perpendicular distances of 1 Oand 2 Ofrom the common chord respectively. Notice an interesting fact : if the length of the common chord ABis 0, it is actually the common tangent to the two circles 1 Cand 2 Cwhich will touch each other externally or internally. Fig - 35 C 2 C 1 P O 1 O 2 C 2 C 1 O 2 O 1 P If the two circles touch each other externally or internally (at), then the common chord – = 0 actually represents the common tangents to the two circles at P S S P 1 2 Now an interesting question arises. Suppose that C 1 and C 2 lie external to each other and do not intersect. What doesS 1 – S 2 = 0 represent in that case? Fig - 36 C 2 C 1 S 1 = 0 S 2 = 0 What does– = 0 represent in this case? Note that no common chord or common tangent can exist. S S 1 2 1 2 0 S S − ·is obviously a straight line. But it is obviously not a common chord or a common tangent since these do not exist in this case. The answer is provided by a slight algebraic manipulation. Let a point 1 1 ( , ) Px ybe such that it satisfies 1 2 0. S S − · Thus, 1 1 1 2 2 1 ( , ) ( , ) S x y S x y · 1 1 1 2 1 1 ( , ) ( , ) S x y S x y ⇒ · 1 2 Assuming both sides are +ve which is true if is external to and. P C C ¹ ¹ ¹ ¹ ' ; ¹ ¹ ¹ ¹ L OCUS L OCUS L OCUS L OCUS L OCUS 41 Mathematics / Circles What does this equation tell us? P is a point such that the lengths of the tangents drawn from it to the two circles are equal. Thus, any point lying on the straight line 1 2 0 S S − ·will posess the property that the tangents drawn from it to the two circles are equal. This line is termed the radical axis of the two circles. Fig - 37 B C 1 For any point on the radical axis, tangents drawn from it to and are of equal length, i.e. = p C C PA PB 1 2 A C 2 This line is the radical axis P It should be obvious that in case of intersecting (or touching) circles, the common chord (or the common tangent) is itself the radical axis. For a situation as in Fig - 37 above, the radical axis exists but no common chord exists, For a circle lying inside another circle, neither the radical axis nor the common chord exist : Fig - 38 C 2 C 1 The radical axis is the same as the common chord. C 2 C 1 X P The radical axis is the same as the common tangent.This should otherwise be obvious also since for any point on this line, the length of tangents to both and is P C C PX 1 2 C 1 The radical axis is the same as the common tangent which is again abvious No common chord or radical axis exist L OCUS L OCUS L OCUS L OCUS L OCUS 42 Mathematics / Circles Show that the radical axis of three circles, whose centres are non-collinear, taken two a + a timeare concurrent. Solution: The significance of the phrase ‘non-collinear’ for the three centres should be clear to you : if the centres are all collinear, the three radical axis will become parallel to each other instead of intersecting. We assume the equations to the three circles to be 1 2 0, 0 S S · ·and 3 0. S ·The three radical axis are therefore 1 1 2 : 0 R S S − · 2 2 3 : 0 R S S − · 3 3 1 : 0 R S S − · Observe that 1 2 3 (1) (1) (1) 0 R R R + + · which implies that 1 2 , R Rand 3 Rare concurrent. The point of concurrency is called the radical centre of the three circles : R 1 R 2 R 3 C 3 C 2 C 1 X R R R X X 1 2 3 , and are the three radical axis, while, the point of concurrency is termed the radical centre. Tangents drawn from to the three circles will be of equal lengths Fig - 39 ____________________________________________________________________________________ Before proceeding we must discuss some properties of two intersecting circles; in particular, we need to understand what we mean by the angle of intersection of two circles. Consider two intersecting circles 1 Cand 2 Cwith radii 1 rand 2 rrespectively and centres at 1 Oand 2 Orespectively: C 2 C 1 C C A B 1 2 and intersect at and. A B Fig - 40 O 2 O 1 Example – 31 L OCUS L OCUS L OCUS L OCUS L OCUS 43 Mathematics / Circles The angle of intersection of the two circles can be defined as the angle between the tangents to the two circles at their point(s) of intersection, which will be the same as the angle between the two radii at the point(s) of intersection. In particular, for example, 1 Cand 2 Cin the figure above intersect at an angle 1 2 . O AO ∠ The most important case we need to consider pertaining to intersecting circles is orthogonal circles, meaning that the angle of intersection of the two circles is a right angle. In that case, the 1 2 . O AO ∠above will become a right - angled one so that 2 2 2 1 2 1 2 O A O A OO + · ...(1) If the two circles and 1 Cand 2 Chave the equations 2 2 2 2 1 1 1 1 1 1 1 1 : 2 2 0 ; S x y g x f y c r g f c + + + + · · + − 2 2 2 2 2 2 2 2 2 2 2 2 : 2 2 0 ; S x y gx f y c r g f c + + + + · · + − then the condition (1) becomes 2 2 2 1 2 1 2 r r OO + · 2 2 2 2 2 2 1 1 1 2 2 2 1 2 1 2 ( ) ( ) g f c g f c g g f f ⇒ + − + + − · − + − 1 2 1 2 1 2 2( ) g g f f c c ⇒ + · + Thus, the condition that the two circles given by 1 Sand 2 Smust satisfy in order to be orthogonal is 1 2 1 2 1 2 2( ) : Orthogonal circles g g f f c c + · + As an exercise, verify that the following pairs of circles intersect orthogonally : (a) 2 2 1 2 2 2 : 6 8 24 0 : 8 6 24 0 S x y x y S x y x y + − − + · + − − + · (b) 2 2 1 2 2 2 : 5 3 7 0 : 8 6 18 0 S x y x y S x y x y + + + + · + − + − · As another example, suppose that we have to find the angle at which 2 2 1 2 2 2 : 6 4 11 0 : 4 6 9 0 S x y x y S x y x y + − + + · + − + + · intersect. We have, 1 2 1 2 2; 2; (3, 2); (2, 3) r r O O · · ≡ − ≡ − L OCUS L OCUS L OCUS L OCUS L OCUS 44 Mathematics / Circles Referring to Fig - 40, let the angle of intersection, 1 2 O AO ∠be θ . By the cosine rule in 1 2 , O AO ∆we have 2 2 2 1 2 1 2 1 2 1 cos 2 2 r r OO r r θ + − · · 4 π θ ⇒ · Thus, the two circles intersect at an angle equal to . 4 π A circle C passes through (1, 1) and cuts the following two circles orthogonally : 2 2 1 2 2 2 : 8 2 16 0 : 4 4 1 0 S x y x y S x y x y + − − + · + − − − · Find the equation of C. Solution: We assume the equation of C to be 2 2 : 2 2 0 S x y gx fy c + + + + · Applying the condition of orthogonality of S with 1 Sand 2 , Swe obtain : with 1 S: 2( 4 ) 16 g f c − − · + 8 2 16 g f c ⇒ + + · − ...(1) with 2 S: 2( 2 2 ) 1 g f c − − · − 4 4 1 g f c ⇒ + + · ...(2) The third condition can be obtained using the fact that C passes through (1, 1): 2 2 1 1 2 (1) 2 (1) 0 g f c + + + + · 2 2 2 g f c ⇒ + + · − ...(3) Solving (1), (2) and (3), we obtain 7 23 , , 5 3 6 g f c · − · · − Thus, the equation of the circle C is 2 2 : 3 3 14 23 15 0 S x y x y + − + − · Example – 32 L OCUS L OCUS L OCUS L OCUS L OCUS 45 Mathematics / Circles Prove that the locus of the centres of the circles cutting two given circles orthogonally is their radical axis. Solution: This assertion means that any circle cutting two given circles orthogonally will have its centre lying on the radical axis of the two given circles. Assume the two fixed given circles to have the following equations: 2 2 1 1 1 1 : 2 2 0 S x y g x f y c + + + + · 2 2 2 2 2 2 : 2 2 0 S x y gx f y c + + + + · Let the variable circle be 2 2 : 2 2 0 S x y gx fy c + + + + ·so that its centre is ( , ) g f − −whose locus we wish to determine. Applying the condition for orthogonality, we obtain 1 1 1 2( ) gg ff c c + · + ...(1) 2 2 2 2( ) gg ff c c + · + ...(2) By (1) – (2), we obtain 1 2 1 2 1 2 2 ( ) 2 ( ) gg g f f f c c − + − · − Using( , ) xyinstead of ( , ), g f − −we obtain the locus as 1 2 1 2 1 2 2 ( ) 2 ( ) ( ) 0 xg g y f f c c − + − + − · which is the same as the equation of the radical axis, i.e. 1 2 0. S S − · We can also prove this assertion of this question in a very straightforward manner using a pure geometric approach. Let C be a circle intersecting the two given circles 1 Cand 2 Corthogonally as show in the figure below: Fig - 41 C 1 C C C OA OA OB OB intersects and orthogonally. Thus, and 1 2 1 2 ⊥ ⊥ A C 2 O O 2 O 1 B C 1 OA O A ⊥and 2 OB O B ⊥implies that OA and OB are simply the tangents drawn from O to 1 Cand 2 C . Since OA and OBare the radii of the same circle C, we have . OA OB · Example – 33 L OCUS L OCUS L OCUS L OCUS L OCUS 46 Mathematics / Circles Thus, O is a point such that tangents drawn from it to the two circles 1 Cand 2 Care equal in length, which implies that O lies on the radical axis of 1 Cand 2 C . From this property, a straightforward corollary follows. For three fixed circles, a circle with centre at the radical centre and radius equal to the length of the tangent from it to any of the circles will intersect all the three circles orthogonally. As an exercise, find the equation to the circle C cutting the following three circles orthogonally : 2 2 1 : 2 3 7 0 C x y x y + − + − · 2 2 2 : 5 5 9 0 C x y x y + + − + · 2 2 3 : 7 9 29 0 C x y x y + + − + · Prove that two circles, both of which pass through the point (0, ) aand (0, ) a −and touch the line, y mx c · +will cut orthogonally if 2 2 2 (2 ) c a m · + Solution: It should be clear that the y-axis is the common chord of the two circles 1 Cand 2 C : C 2 C 1 y C C C C x - axis is the common chord of and Also, the centres of both and with lie on the - axis 1 2 1 2 Fig - 42 y (0 ) 1 ,a (0 ) ,-a x Let the centres of 1 Cand 2 Cbe 1 ( , 0) g −and 2 ( , 0) g −so that their radii become 2 2 1 g a + and 2 2 2 g a +respectively. Their equations then become : 2 2 2 1 1 : 2 0 C x y g x a + + − · 2 2 2 2 2 : 2 0 C x y gx a + + − · Example – 34 L OCUS L OCUS L OCUS L OCUS L OCUS 47 Mathematics / Circles The liney mx c · +touches both 1 Cand 2 Cso that the perpendicular distance of the centres of 1 C and 2 Cfrom this line must be respectively equal to their radii. Thus, we obtain 1 2 2 1 2 1 mg c g a m − · + + and 2 2 2 2 2 1 mg c g a m − · + + 2 2 2 2 1 1 2 (1 ) 0 g mcg a m c ⇒ + + + − · and 2 2 2 2 2 2 2 (1 ) 0 g mcg a m c + + + − · Thus, 1 gand 2 gare the roots of the equation 2 2 2 2 2 (1 ) 0 g mcg a m c + + + − · so that 2 2 2 1 2 (1 ) g g a m c · + − Finally, 1 Cand 2 Care orthogonal if the condition 1 2 1 2 1 2 2( ) g g f f c c + · + is satisfied, i.e. 2 2 2 2 2{( (1 ) ) (0)(0)} 2 a m c a + − + · − 2 2 2 (2 ) c a m ⇒ · + This is the required relation for 1 Cand 2 Cto be orthogonal. Consider the following two circles: 2 2 1 2 2 2 : 16 0 : 8 12 16 0 S x y S x y x y + − · + − − + · Circles are drawn which are orthogonal to both these circles. Tangents are drawn from the centre of the variable circle to 1 S . Find the mid locus of the mid-point of the chord of contact so formed. Solution: The centre of the variable circle, say 1 1 ( , ), x ylies on the radical axis of the two given circles, i.e. on 1 2 0 S S − ·by the result obtained in Example - 33. Example – 35 L OCUS L OCUS L OCUS L OCUS L OCUS 48 Mathematics / Circles Thus, 1 1 ( , ) x ymust satisfy 1 2 0 S S − ·or8 12 32 0 x y + − · 1 1 8 12 32 0 x y ⇒ + − · ...(1) From 1 1 ( , ), x ytwo tangents are drawn to 1 . SThe equation of the chord of contact is therefore 1 1 ( , ) 0 Tx y · 1 1 16 xx yy ⇒ + · ...(2) Let the mid-point of the chord of contact so formed be( , ). Mh kWe can also write the equation of the same chord ofcontact using the equation we derived for a chord bisected at a given point. Thus, the chord of contact can also be represented by the equation ( , ) ( , ) Th k Sh k · 2 2 hx ky h k ⇒ + · + ...(3) Since (2) and (3) are the same lines, we have 1 1 2 2 16 x y h k h k · · + 1 1 2 2 2 2 16 16 ; h k x y h k h k ⇒ · · + + ...(4) Using (4) in (1), we finally obtain a relation in ( , ) h k : 1 1 8 12 32 0 x y + − · 2 2 2 2 128 192 32 h k h k h k ⇒ + · + + 2 2 4 6 0 h k h k ⇒ + − − · Using ( , ) xyinstead of ( , ) h kwe obtain the required locus as 2 2 4 6 0 x y x y + − − · L OCUS L OCUS L OCUS L OCUS L OCUS 49 Mathematics / Circles Q. 1 Find the equation of the circle which intersects 2 2 6 4 3 0 x y x y + − + − ·orthogonally, passes through (3, 0) and touches the y-axis. Q. 2 At what angle do the circles 2 2 1 2 2 2 : 4 6 11 0 : 2 8 13 0 S x y x y S x y x y + − + + · + − + + · intersect ? Q. 3 Find the co-ordinates of the point from which the tangents drawn to the following three circles are of equal lengths : 2 2 1 2 2 2 2 2 3 : 3 3 4 6 1 0 : 2 2 3 2 4 0 : 2 2 1 0 S x y x y S x y x y S x y x y + + − − · + − − − · + − + − · Q. 4 Find the locus of the centre of the circle passing through ( , ) a band orthogonal to 2 2 2 . x y k + · Q. 5 Let 1 1 ( , ) A a band 2 2 ( , ) B a bbe two fixed points and O be the origin. Circles are drawn on OA and OB as diameters. Prove that the length of the common chord is 1 2 2 1 a b a b AB − . TRY YOURSELF - III L OCUS L OCUS L OCUS L OCUS L OCUS 50 Mathematics / Circles By a family of circles, we will mean a set of circles satisfying some given property (or properties). For example, the family of circles with each circle having its centre lying in the first quadrant and touching both the co-ordinate axes can be represented by the equation 2 2 2 ( ) ( ) x a y a a − + − · ...(1) where a is a positive real number. The important point to observe is that a is a variable here. As we vary a, we get different circles belonging to this family, but due to the constraint imposed by (1), all circles of this family satisfy the specified property. y x Some members of the family of circles given by (1). The centre of a circle is ( ,) and its radius is, where is a real positive variable. As is varied, we obtain different members in this family aa a a a Fig - 43 We intend to discuss in this section certain families that are of significant importance. In all cases, the family will be represented by an equation containing a real variable, which when varied will give rise to different members of this family. TYPE 1: FAMILY OF CIRCLES PASSING THROUGH THE INTERSECTION POINTS OF A GIVEN CIRCLE AND A GIVEN LINE We are given a fixed circle and a fixed line with equations 0 S · and 0 L · respectively. We want to find out the equation of the family of the circles passing through the points of intersection of 0 S · and 0 L · . The circles = 0 and = 0 are fixed. The dotted circles represent some of the members of the family of circles passing through the intersection points and of= 0 and = 0 S L A B S L L = 0 B S = 0 A Fig - 44 Sec t i on-4 FAMI LY OF CI RCLES L OCUS L OCUS L OCUS L OCUS L OCUS 51 Mathematics / Circles By now, it should be apparent to you how to specify the equation of this family in terms of a real variable. Any circle F belonging to this family can be written as 0 F S L λ ≡ + ·where λ ∈! . The truth of this assertion can be easily verified. F as defined from this equation is definitely a circle since it satisfies the required constraints for its equation to be a circle. To see this, let 2 2 : 2 2 0 S x y gx fy c + + + + · : 0 L px qy r + + · We then have : 0 F S L λ + · 2 2 (2 ) (2 ) 0 x y g px f qy c r ⇒ + + + + + + + · λ λ λ which is definitely the equation of a circle. To see that F passes through A and B, note that since A and B satisfy both the equations 0 S · and0, L ·they will obviously also satisfy the equation 0. S L λ + · Thus, 0 F S L λ ≡ + · is the required family of circles. As we vary λ , we will get different circles belonging to this family. In particular, note that for0, λ ·we get the circle 0 S · itself, while in the limit, λ → ∞we get the ‘circle’0, L ·which is actually a line but can be considered a circle with centre lying at infinity and an infinitely large radius. Write the equation for (a) the family of circles passing through 1 1 ( , ) x yand 2 2 ( , ) x y (b) the family of circles touching the line 0 L · at 1 1 ( , ) x y Solution (a) Till now we have seen how to write the equation for a family of circles passing through the intersection points of a circle and a line. Thus, in the current example, we first define a fixed circle passing through 1 1 ( , ) x yand 2 2 ( , ) x yand the line through these two points. The fixed circle S can easily be taken to be the one with 1 1 ( , ) x yand 2 2 ( , ) x yas the end points of a diameter : 1 2 1 2 : ( )( ) ( )( ) 0 S x x x x y y y y − − + − − · The fixed line can be written using the two-point form : 1 1 1 1 2 1 2 1 2 2 1 : : 1 0 1 x y y y x x L L x y y y x x x y − − · ⇒ · − − Example – 36 L OCUS L OCUS L OCUS L OCUS L OCUS 52 Mathematics / Circles Given two fixed points ( ,) and ( ,), we define the fixed circle as the one with these points being the end-points of a diameter.The fixed line is simply the line passing through these two points. x y x y 1 1 2 2 S = 0 L = 0 ( ,) x y 2 2 ( ,) x y 1 1 Fig - 45 We can now write the required family of circles as : : 0 F S L λ + · 1 2 2 2 ( )( ) ( )( ) 0 x x x x y y y y L ⇒ − − + − − + · λ ...(1) where 0 L · represents the line through the two given points (b) We now want the family of circles to be such that each circle touches 0 L · at 1 1 ( , ) x y . We could evaluate it by letting 2 1 x x →and 2 1 y y →in (1). In this limit, 0 L · will simply become the tangent to any member of the family F given by (1). (Convince yourself about this point. Thus, the required family is 2 2 1 1 : ( ) ( ) 0 F x x y y L λ ′ − + − + · TYPE 2: FAMILY OF CIRCLES TOUCHING A GIVEN CIRCLE AT A GIVEN POINT Let the equation of the fixed circle be 2 2 : 2 2 0 S x y gx fy c + + + + · andlet there be a point 1 1 ( , ) Px ylying on this circle. We wish to determine the equation of the family of circles touching S at P. The dotted circles are some of the members of the family of circles in which each circle touches atS P S = 0 P ( ,) x y 1 1 Fig - 46 L OCUS L OCUS L OCUS L OCUS L OCUS 53 Mathematics / Circles We can write the equation of the tangent to 0 S · at P as 1 1 1 1 : ( ) ( ) 0 T xx yy gx x f y y c + + + + + + · Once we have a circle ( 0) S ·and a line ( 0) T ·intersecting or touching the circle, we can write the equation of the family of circles passing through the point (s) of intersection of the circle and the line, using the result derived in the last article. Thus, the required family can be represent as : 0 F S T λ + · 2 2 1 1 1 1 : 2 2 ( ( ) ( ) 0 F x y gx fy c xx yy gx x f y y c λ ⇒ + + + + + + + + + + + · 2 2 1 1 1 1 : (2 ) (2 ) 0 F x y g x gx f y f y c gx fy c ⇒ + + + + + + + + + + + · λ λ λ λ λ λ λ As we vary, λwe will obtain different members belonging to this family. TYPE 3: FAMILY OF CIRCLES PASSING THROUGH THE INTERSECTION POINT(S) OF TWO GIVEN CIRCLES Let the two fixed circles be 2 2 1 1 1 1 : 2 2 0 S x y g x f y c + + + + · 2 2 2 2 2 2 : 2 2 0 S x y gx f y c + + + + · and their points of intersection be 1 1 ( , ) A x yand 2 2 ( , ). Bx yIn case the two circles touch each other, A and B will be the same. We wish to determine the family of circles passing through A and B. The dotted circles represent some members of the family of circles in which each member passes through andA B Fig - 47 S= 2 0B S= 1 0 A You might be able to extrapolate from the last few cases that the equation representing this family will be 1 2 : 0 F S S λ + · L OCUS L OCUS L OCUS L OCUS L OCUS 54 Mathematics / Circles You can verify this by writing the equation for F in standard form and observing that it does indeed represent a circle. Also, since (the co-ordinates of) A and B satisfy both 1 0 S·and 2 0, S ·they have to satisfy the equation for F. Note one important point: λ cannot be equal to –1 otherwise F will become the common chord of 1 0 S·and 2 0 S ·instead of representing a circle. Find the equation of the circle which passes through the points of intersection of the circles 2 2 1 : 6 2 4 0 S x y x y + − + + · 2 2 2 : 2 4 6 0 S x y x y + + − − · and whose centre lies on the line y = x. Solution: Let the required equation be S = 0. Then, by the previous article, we can find someλ∈! and 1 λ ≠ − such that 1 2 0 S S S ≡ + λ · 2 2 (1 ) (1 ) (2 6) (2 4 ) 4 6 0 S x y x y ⇒ ≡ + λ + + λ + λ − + − λ + − λ ·...(1) The centre of S from this equation comes out to be ( 3) (1 2 ) centre , 1 1 λ − − λ ¹ ¹ ≡ − − ' ; + λ + λ ¹ ¹ Since the centre lies on the line y = x, we have ( 3) (1 2 ) 1 1 λ − − λ − · − + λ + λ 4 3 ⇒ λ · We now substitute this value back in (1) to obtain the equation for S: 2 2 7 7 10 10 : 4 0 3 3 3 3 S x y x y + − − − · 2 2 : 10 10 12 0 S x y x y ⇒ + − − − · A family of circles passing through the points A (3, 7) and B (6, 5) cuts the circle 2 2 4 6 3 0. x y x y + − − − ·Show that the common chord of the fixed circle and the variable circle(belonging to the family) will always pass through a fixed point. Find that point. Example – 37 Example – 38 L OCUS L OCUS L OCUS L OCUS L OCUS 55 Mathematics / Circles Solution: We can write the equation to the specified family of circles by first writing the equation L of the line AB: 7 2 : 3 3 y L x − · − − : 2 3 27 L x y ⇒ + · The required family can now be written as : ( 3)( 6) ( 7)( 5) 0 F x x y y L − − + − − + λ ·where λ∈! 2 2 : (2 9) (3 12) (53 27 ) 0 F x y x y ⇒ + + λ − + λ − + − λ · The common chord of F and the given circle S is : 0 S F − · (5 2 ) (6 3 ) (27 56) 0 x y ⇒ − λ + − λ + λ − · (5 6 56) (2 3 27) 0 x y x y ⇒ + − − λ + − · 1 2 0 L L ⇒ +µ · That the common chord can be written like 1 2 0 L L +µ ·implies that it will always pass through the intersection point of L 1 and L 2 , what ever the value of µ may be. This intersection point can be obtained (by simultaneously solving L 1 and L 2 ) to be 23 2, 3 ¸ _ ¸ , . Find the equation of the circle which touches the line0 x y − ·at the origin and bisects the circumference of the circle 2 2 2 3 0 x y y + + − · . Solution: In example -36, we evaluated the equation of the family of circles all touching a given line at a given point. Here, the given line is0 x y − ·and the given point is (0, 0). Thus, the equation of the family is 2 2 : ( 0) ( 0) ( ) 0 F x y x y − + − + λ − · 2 2 : 0 F x y x y ⇒ + + λ −λ · ...(1) We need to find the value of λ for which the circle in (1) bisects the circumference of the given circle 2 2 : 2 3 0, S x y y + + − ·which means that the common chord of the required circle and S will the diameter of S. Example – 39 L OCUS L OCUS L OCUS L OCUS L OCUS 56 Mathematics / Circles A B The variable circle S = 0 If the variable circle bisects the circumference of, this mean that the common chord must be the diameter of. S AB S Fig - 48 The common chord AB is 0 F S − · ( 2) 3 0 x y ⇒ λ −λ + + · Since this is the diameter of S, the centre of S, i.e. (0, –1), must lie on it (satisfy its equation). Thus, we obtain λ as (0) ( 2)( 1) 3 0 λ −λ + − + · 5 ⇒ λ · − Finally, we substitute this value of λ back in (1) to get the required equation of the circle as 2 2 5 5 0. x y x y + − + · Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4 3 10, x y + ·find the equation of the circles. Solution: The following diagram explains the situation better: 4+ 3= 10 x y The two circles touch each other at (1, 2). Fig - 49 C 2 C 1 S 2 = 0 (1, 2) S 1 = 0 We describe here two alternatives that can be used to solve this question Example – 40 L OCUS L OCUS L OCUS L OCUS L OCUS 57 Mathematics / Circles Alternative - 1: Find the centres of the two circles The slope of the common tangent is 4 . 3 − Therefore, the slope of C 1 C 2 is 3 , 4 i.e. 3 tan 4 m · θ · 3 sin 5 ⇒ θ · and 4 cos 5 θ · Using the polar form, we can write the co-ordinates of any point on the line C 1 C 2 : 1 2 cos sin x y r − − · · θ θ 1 cos , 2 sin x r y r ⇒ · + θ · + θ Substituting5 r · tgives the two centres as should be apparent from the figure. Thus, the two centres are 1 (5, 5) C≡and 2 ( 3, 1) C≡− − Thus, the two equations are 2 2 1 : 10 10 25 0 S x y x y + − − + · 2 2 2 : 6 2 15 0 S x y x y + + + − · Alternative - 2: Use a family of circles approache The family of circles touching L = 0 at 1 1 ( , ) x ycan be written, as described earlier, as 2 2 1 1 ( ) ( ) 0 x x y y L − + − + λ · In the current case, this becomes 2 2 ( 1) ( 2) (4 3 10) 0 x y x y − + − + λ + − · 2 2 (4 2) (3 4) (5 10 ) 0 x y x y ⇒ + + λ − + λ − + − λ · ...(1) The radius of the required circle is 5. Thus, 2 2 2 3 4 (2 1) (5 10 ) 5 2 λ − ¸ _ λ − + − − λ · ¸ , This is a quadratic in λ which gives two values of λ : 2 λ · t Using these values in (1), we obtained the two required circles. L OCUS L OCUS L OCUS L OCUS L OCUS 58 Mathematics / Circles Q. 1 Find the equations of the circles with radius 4 and passing through the points of intersection of 2 2 1 : 2 4 4 0 S x y x y + − − − · 2 2 2 : 10 12 40 0 S x y x y + − − + · Q. 2 Find the equation of the circle whose diameter is the common chord of the circles 2 2 2 ( ) x a y a − + · and 2 2 2 ( ) x y b b + − · . Q. 3 Find the equation of the circle passing through (2, 1) and touching the line2 1 x y + ·at (3, –1). Q. 4 Tangents are drawn from the origin to 2 2 6 4 12 0. x y x y + + + − ·Find the equation of the circle passing through the points of contact of these tangents and the origin. Q. 5 Find the equation of a circle which touches the line5 x y + ·at the points (–2, 7) and cuts the circle 2 2 4 6 9 0 x y x y + + − + ·orthogonally. TRY YOURSELF - IV L OCUS L OCUS L OCUS L OCUS L OCUS 59 Mathematics / Circles The circle 2 2 4 4 4 0 x y x y + − − + ·is inscribed in a triangle which has two of its sides along the co-ordinate axes. If the locus of the circumcentre of the triangle is 2 2 0 x y xy k x y + − + + · find the value of k. Solution : The situation is described clearly in the figure below: (2,2) y x The circle = 0 is fixed. The line is variable, intersecting the axis in( , 0) and (0,) respectively. We are concerned with the locus of the circumcentre of. S A a B b OAB ∆ Fig - 50 B (0 ) , b S= 0 0 ( 0) a, A L The equation of L is, using the intercept form, : 1 x y L a b + · The distance of the centre of S, i.e. (2, 2) from L must equal the radius of S which is 2. Thus, 2 2 2 2 1 2 1 1 a b a b + − · + We now use the fact that (2, 2) is negative since (2, 2) and the origin lie on the same side of and (0, 0) is negative L L L ¹ ¹ ¹ ¹ ¹ ¹ ' ; ¹ ¹ ¹ ¹ ¹ ¹ 2 2 2 2 2 0 a b ab a b ⇒ + − + + · ...(1) From pure geometric considerations, the circumcentre C ofOAB ∆lies on AB and is in fact, the mid-point of AB. Thus, , 2 2 a b C ¸ _ ≡ ¸ , SOLVEDEXAMPLES Example – 01 L OCUS L OCUS L OCUS L OCUS L OCUS 60 Mathematics / Circles Slightly manipulating (1), we obtains 2 2 0 2 2 2 2 2 2 a b a b a b ¸ _¸ _ ¸ _ ¸ _ + − + + · ¸ ,¸ , ¸ , ¸ , ...(2) The locus of, 2 2 a b ¸ _ ¸ , is given by (2). Using ( , ) xyinstead of, , 2 2 a b ¸ _ ¸ , we obtain 2 2 0 x y xy x y + − + + · ...(3) Upon comparing (3) with the locus specified in the question, we obtain 1. k · Consider a curve 2 2 2 1 ax hxy by + + ·and a point P not on the curve. A line drawn from P intersects the curve at points Q and R. If the productPQPR ⋅is independent of the slope of the line, then show that the curve is a circle. Solution : Since distances are involved from a fixed point, it would be a good idea to use the polar form of the line to write the co-ordinates of Q and R. Let P be 1 1 ( , ) x yand let θ denote the slope of the variable line. For any point ( , ) xylying on this line at a distance r from P, we have 1 1 cos sin x x y y r θ θ − − · · 1 1 cos ; sin x x r y y r θ θ ⇒ · + · + If ( , ) xylies on the given curve, it must satisfy the equation of the curve : 2 2 1 1 1 1 ( cos ) ( sin ) 2 ( cos )( sin ) 1 ax r by r hx r y r θ θ θ θ + + + + + + · 2 2 2 1 1 1 { cos sin sin 2 } 2{ cos sin ( cos a b h r ax by hx θ θ θ θ θ θ ⇒ + + + + + 2 2 1 1 1 1 1 sin )} 2 1 0 y r ax by hx y θ + + + + − · ...(1) This quadratic has two roots in r, say 1 rand 2 r , which will actually correspond toPQ and PR since Q and R lie on the curve. Thus,PQPR ⋅being independent of θ means that 1 2 r rfor (1) is independent of θ i.e. 2 2 1 1 1 1 2 2 2 1 cos θ sin θ sin 2θ ax by hx y a b h + + − + + is independent of θ 2 2 1 1 1 1 2 1 cos 2θ sin 2θ 2 2 ax by hx y a b a b h + + − ⇒ + − ¸ _ ¸ _ + + ¸ , ¸ , is independent of θ Example – 02 L OCUS L OCUS L OCUS L OCUS L OCUS 61 Mathematics / Circles 2 2 1 cos 2 1 cos 2 The denominator was obtained in this form by writingcosas andsinas 2 2 + − ¹ ¹ ' ; ¹ ¹ θ θ θ θ 2 2 1 1 1 1 2 2 2 1 {sin(2 } 2 2 ax by hx y a b a b h θ φ + + − ⇒ + − ¸ _ ¸ _ + + + ¸ , ¸ , is independent of θ The denominator was obtained by combining the two trignometric terms. tan is 2 a b h ¹ − ¹ ¸ _ ' ; ¸ , ¹ ¹ φ ⇒ This is only possible when 2 2 0 2 a b h − ¸ _ + · ¸ , ⇒ a b · and 0 h · Thus, the equation of the given curve reduces to 2 2 1 ax ay + · which is clearly the equation of a circle. The equation of a circle is: 2 ( ) (2 ) 0 S xx a y y b − + − ·where, 0. a b ≠Find the condition on a and b if two chords each bisected by the x-axis, can be drawn to the circle from the point, . 2 b P a ¸ _ ¸ , Solution: Observe that the centre of S is , 2 4 a b ¸ _ ¸ , and P lies on the circle: Fig - 51 The centre of the circle is (); S y x the point lies on the circle. In fact, is a diameter of the circle P OP a 2 b 4 , a 2 b 4 , ( ) a b 2 , () O P Example – 03 L OCUS L OCUS L OCUS L OCUS L OCUS 62 Mathematics / Circles Let us consider a chordPQof the circle bisected at the x-axis, say, at the point ( , 0). hWe can write the equation ofPQ as : ( , 0) ( , 0) Th Sh · where( , ) Sxyis 2 2 0 2 b x y ax y + − − · 2 ( ) ( 0) 2 4 a b hx x h y h ah ⇒ − + − + · − 2 0 2 4 2 a b ah h x y h ¸ _ ¸ _ ⇒ − − + − · ¸ , ¸ , ...(1) Since this chord passes through, , 2 b P a ¸ _ ¸ , the co-ordinates of (P) must satisfy (1). Thus, 2 0 2 4 2 2 a b b ah h a h ¸ _ ¸ _ − − + − · ¸ , ¸ , 2 2 2 3 0 2 2 8 ah a b h ⇒ − − − · 2 2 2 3 0 2 2 8 ah a b h ⇒ − + + · ...(2) We want to have two such possible chords PQ bisected at the x-axis. Thus, we must have two distinct values of h which can happen if the discriminant of (2) is positive. Thus, 2 2 2 9 4 4 2 8 a a b ¸ _ > + ¸ , 2 2 2 a b ⇒ > This is the required condition that a and b must satisfy. Let 1 Tand 2 Tbe two tangents drawn from (–2, 0) to the circle 2 2 1. x y + ·Determine the circles touching C and having 1 Tand 2 Tas their pair of tangents. Also find all the possible pair wise common tangents to these circles. Example – 04 L OCUS L OCUS L OCUS L OCUS L OCUS 63 Mathematics / Circles Solution: Note that two such circles can be drawn, as shown in the figure below : R y x C C x x 1 2 1 2 and are the two possible circles, with centres ( , 0) and ( , 0) − Fig - 52 B x, ( 0) 2 Q P S ( 2 0) - , x+ y= 1 22 ( ) 1 Centre at ,0 x − C 1 O A R' C 2 Let the radii of 1 Cand 2 Cbe 1 rand 2 rrespectively. We can evaluate the coordinates of both A and B very easily from geometrical considerations. We have ASP OSQ BSR ∆ ∆ ∆ ∼ ∼ AS OS BS AP OQ BR ⇒ · · Now, 1 1 2 2 2 , , 2, 1, 2 , AS x AP r OS OQ BS x BR r · − · · · · + · 1 2 1 2 2 2 2 1 x x r r − + ⇒ · · Also, note that 1 1 1 x r · +and 2 2 1 . x r · +Thus, 1 2 1 2 2 2 2 1 1 x x x x − + ⇒ · · − − 1 4 3 x ⇒ · and 2 4 x · 1 1 3 r ⇒ · and 2 3 r · L OCUS L OCUS L OCUS L OCUS L OCUS 64 Mathematics / Circles Thus, the equations to 1 Cand 2 Care 2 2 2 1 4 1 : 3 3 C x y ¸ _ ¸ _ + + · ¸ , ¸ , 2 2 2 2 : ( 4) 3 C x y − + · Now we determine all the possible common tangents to 1 Cand 2 C . Two such possible tangents are simply SR and SR′ which both touch 1 Cand 2 Con the same side. There will be two other possible common tangents, each having A and B on the opposite side of it as shown in the figure below : y x Thesetangents are termed the transverse tangents to and. C C 1 2 Fig - 53 P A Q C 2 B θ C 1 X Let X have the co-ordinates ( , 0). xNote that. APX BQX ∆ ∆ ∼ Thus, AX BX AP BQ · 4 4 3 1/ 3 3 x x + − ⇒ · 4 5 x ⇒ · − Also, 1 3 5 sin 4 5 4 3 8 AP AX θ · · · − + 2 2 5 5 tan 39 8 5 θ ⇒ · · − Thus, the two transverse tangents pass through 4 , 0 5 − ¸ _ ¸ , and have slopes 5 . 39 tTheir equations are therefore 5 4 5 39 y x ¸ _ · t + ¸ , L OCUS L OCUS L OCUS L OCUS L OCUS 65 Mathematics / Circles Find the values of a for which the line0 x y + ·bisects two chords drawn from a point 1 2 1 2 , 2 2 a a P ¸ _ + − ¸ , to the circle 2 2 1 2 1 2 : 0 2 2 a a S x y x y ¸ _ ¸ _ + − + − − · ¸ , ¸ , . Solution: This question is somewhat similar to Example -3. Note that the point P lies on S. Since the mid-point M of the chord(s) drawn from P lies on0, x y + ·we can assume its co-ordinates to be( , ). h h −The equation of the chord of S bisected at M then becomes: ( , ) ( , ) Th h Sh h − · − 2 ( ) ( ) 2 2 2 hx hy x h y h h h h λ λ λ λ ⇒ − − + − − · − + ...(1) where the substitution 1 2 1 2 , 2 2 a a λ λ + − · ·has been done for convenience. Since the chord (1) passes through( , ), P λλits co-ordinates must satisfy (1). Thus, 2 ( ) ( ) 2 2 2 h h h h h h h λ λ λ λ λ λ λ λ − − + − − · − + 2 2 2 4 3( ) ( ) 0 h h λ λ λ λ ⇒ − − + + · ...(2) Since we want two possible chords to exist satisfying the given property, the quadratic in (2) must yield distinct values of h. Thus, its discriminant must be positive : 2 2 2 9( ) 16( ) λ λ λ λ − > + Substituting the values of λ and λ and simplifying, we obtain 2 4 0 a − > ( , 2) (2, ) a ⇒ ∈−∞ − ∪ ∞ These are the required values of a. If 1 0 S·and 2 0 S ·are circles having radii 1 rand 2 rrespectively, prove that the circles 1 2 1 2 : 0 S S S r r + ·and 1 2 1 2 ' : 0 S S S r r − ·intersect orthogonally. Example – 05 Example – 06 L OCUS L OCUS L OCUS L OCUS L OCUS 66 Mathematics / Circles Solution: Let the equation of the two given circles be 2 2 2 2 2 1 1 1 1 1 1 1 1 : 2 2 0 ; S x y g x f y c r g f c + + + + · · + − 2 2 2 2 2 2 2 2 2 2 2 2 2 : 2 2 0 ; S x y gx f y c r g f c + + + + · · + − Then, the equation of S and S ' are 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 : ( ) 2 2 0 g g f f c c S x y x y r r r r r r r r ¸ _ ¸ _ ¸ _ ¸ _ + + + + + + + + · ¸ , ¸ , ¸ , ¸ , 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 : ( ) 2 2 0 g g f f c c S x y x y r r r r r r r r ¸ _ ¸ _ ¸ _ ¸ _ ′ + − + − + − + − · ¸ , ¸ , ¸ , ¸ , WenowchecktheorthogonalityconditionforSandS' ,i.e.wecheckwhether ' ' ' 2( ) S S S S S S gg f f c c + · + Thus, 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 ' ' 1 2 1 2 1 2 1 2 2( ) 2 2 1 1 1 1 1 1 1 1 S S S S g g g g f f f f r r r r r r r r gg f f r r r r r r r r ¸ _¸ _ ¸ _¸ _ + − + − ¸ ,¸ , ¸ ,¸ , + · + ¸ _¸ _ ¸ _¸ _ + − + − ¸ ,¸ , ¸ ,¸ , 2 2 2 2 1 1 2 2 2 2 1 2 2 2 1 2 2 1 1 g f g f r r r r ¸ _ + + − ¸ , · ¸ _ − ¸ , 1 2 2 2 1 2 2 2 1 2 2 1 1 c c r r r r ¸ _ − ¸ , · ¸ _ − ¸ , ...(1) whereas 1 2 1 2 1 2 1 2 ' 1 2 1 2 1 1 1 1 S S c c c c r r r r c c r r r r + − + · + + − 1 2 2 2 1 2 2 2 1 2 2 1 1 c c r r r r ¸ _ − ¸ , · ¸ _ − ¸ , ... (2) The equality of (1) and (2) implies that the two circles are orthogonal. L OCUS L OCUS L OCUS L OCUS L OCUS 67 Mathematics / Circles Let 1 0 S·and 2 0 S ·be two circles with radii 1 rand 2 rrespectively. Find the locus of the point P at which the two circles subtend equal angles. Solution: Consider 1 0 S·which subtends an angle α(say) at( , ) P h k ( ) -g,-f C 1 P hk ( ,) S 1 = 0 T Fig - 54 We have, 1 1 CT r ·and 1 ( , ) PT S h k · Thus, in 1 , C PT ∆we have 1 1 tan 2 ( , ) r S h k · α ...(1) We can write an exactly analogous equation for 2 Swhich subtends the same angle αat P: 2 2 tan 2 ( , ) r S h k α · ...(2) From (1) and (2), we have 1 2 1 2 ( , ) ( , ) r r S h k S h k · Using ( , ) xyinstead of ( , ), h kwe obtain the required locus as 1 2 2 2 1 2 ( , ) ( , ) S xy S xy r r · Circles are drawn passing through the origin O to intersect the co-ordinate axes at points P and Q such that mOP nOQ ⋅ + ⋅is a constant. Show that the circles pass through a fixed point other than the origin. Example – 07 Example – 08 L OCUS L OCUS L OCUS L OCUS L OCUS 68 Mathematics / Circles Solution: Consider one such circle as shown in the figure below: Fig - 55 y x Given that + = (a constant) we need to show that will always pass through a fixed point mOP nOQ k S O C S q (0, ) Q P p, ( 0) We have OP p ·and. OQ q ·Thus, mp nq k + · ...(1) Now, observe that the centre C of the circle will be , 2 2 p q ¸ _ ¸ , and its radius will be OC. Thus, the equation of the circle is 2 2 2 2 2 2 4 4 p q p q x y ¸ _ ¸ _ − + − · + ¸ , ¸ , 2 2 0 x y px qy ⇒ + − − · ...(2) From (1), we can find q in terms of p and substitute in (2) so that (2) becomes 2 2 0 mp k x y px y n − ¸ _ + − + · ¸ , 2 2 { ( ) } { } 0 nx y ky p nx my ⇒ + − + − + · We see that the variable circle can be written as 0 0 0, S pL p + · ∈! where 2 2 0 ( ) 0 S nx y ky ≡ + − ·and 0 0 L nx my ≡ − + · . Thus, the variable circle will always pass through the two intersection points of S 0 and L 0 , one of them obviously being the origin. The other fixed point can be obtained by simultaneously solving the equations of 0 Sand 0 L to be 2 2 2 2 , mk nk m n m n ¸ _ + + ¸ , L OCUS L OCUS L OCUS L OCUS L OCUS 69 Mathematics / Circles The two curves 2 2 1 : 2 2 2 0 S ax hxy by gx fy c + + − − + · 2 2 2 : 2 ( ) 2 2 0 S a x hxy a a by g x f y c ′ ′ ′ ′ − + + − − − + · intersect at four concyclic points, , ABCand. DLet P be the point , . g g f f a a a a ′ ′ + + ¸ _ ′ ′ + + ¸ , Find the value of 2 2 2 2 . PA PB PC PD + + Solution: From what we’ve learnt in this chapter, we can write the equation of an arbitrary curve S passing through the intersection points of two given curves, here 1 Sand 2 S , as 1 2 0 S S S λ ≡ + ·where λ ∈! Here, 2 2 : ( ) 2 (1 ) ( ( )) 2 ( ) S a a x h xy b a a b y xg g ′ ′ ′ + + − + + + − − + λ λ λ λ 2 ( ) (1 ) 0 y f f c λ ′ − + + + ·..(1) Since this curve S passes through the four concyclic points, , , ABC Dits equation in (1) must be thatof a circle. Thus, we impose the necessary constraints : 2 2 Coeff. ofCoeff. of( ) x y a a b a a b ′ ′ · ⇒ + · + + − λ λ 1 λ ⇒ · Coeff. of0 1 xy · ⇒ · λ Thus, for1, λ ·the equation in (1) represents a circle. This equation now becomes : 2 2 : ( ) ( ) 2( ) 2( ) 2 0 S a a x a a y g g x f f y c ′ ′ ′ ′ + + + − + − + + · The centre of S therefore is , . g g f f a a a a ′ ′ + + ¸ _ ′ ′ + + ¸ , But this is the same as the point P!. Thus, P is the centre of the circle passing through, , , ABC D so that PA PB PC PD · · · Thus, we simply have 2 2 2 2 3 PA PB PC PD + + · Example – 09 L OCUS L OCUS L OCUS L OCUS L OCUS 70 Mathematics / Circles Let x, y and z be the lengths of the perpendiculars dropped from the circumcentre in a triangle ABC to the sides BC, CA, and ABrespectively. The lengths of BC, CAand AB are a, b and c respectively. Show that 4 a b c abc x y z xyz + + · Solution: We can attempt this question through a co-ordinate approach but the nature of the problem posed suggests that a plane geometric approach would be better since no equations need to be determined here. Refer to the following diagramwhich shows the situation clearly : The circumcentre is. All lengths have been marked. Note that = 2= 2 = O COB A A θ ⇒θ Fig - 56 B a c b O P x θ z y A C In, OPB ∆we have tanθ tan 2 a A x · ∠ · ...(1) Similarly,tan 2 b B y ∠ ·andtan 2 C C z ∠ · ...(2) Now, the angles, A B ∠ ∠and C ∠ are those of a triangle so that A B C π ∠ +∠ +∠ · tan( ) tan( ) A B C π ⇒ ∠ + ∠ · − ∠ tan C · − ∠ tan tan tan 1 tan tan A B C A B ∠ + ∠ ⇒ · − ∠ − ∠ ∠ tan tan tan tan tan tan A B C A B C ⇒ ∠+ ∠ + ∠ · ∠ ∠ ∠ Using (1) and (2) in the relation above, we obtain the required relation. Example – 10 L OCUS L OCUS L OCUS L OCUS L OCUS 71 Mathematics / Circles Let( , 0) A abe a point on the circle 2 2 2 . x y a + ·Through another point(0, ), B bchords are drawn to meet the circle at points Q and R. Find the locus of the centroid of. AQR ∆ Solution : The following diagram shows an example of the situation described : The points and are fixed. The line is of variable slope. We need to determine the locus of the centroid of A B BQR AQR ∆ Fig - 57 Q R A a,( 0) B , b (0 ) The equation ofBQRcan be written asy mx b · +where m is variable. The intersection points of BQRwith the circle (Q and R) are given by simultaneously solving the system of equations 2 2 2 x y a + · y mx b · + Thus, 2 2 2 ( ) x mx b a + + · 2 2 2 2 (1 ) 2 0 m x mbx b a ⇒ + + + − · ...(1) If we assume the co-ordinates of Q and R to 1 1 ( , ) x yand 2 2 ( ), xywe have from (1) 1 2 2 2 1 mb x x m + · − + Thus, 2 1 2 1 2 2 2 2 2 ( ) 2 2 1 1 m b b y y mx x b b m m + · + + · − + · + + Let the centroid ofAQR ∆be ( , ). h kWe have 1 2 1 2 0 , 3 3 a x x y y h k + + + + · · 1 2 1 2 3 , 3 x x h a y y k ⇒ + · − + · 2 2 2 2 3 , 3 1 1 mb b h a k m m ⇒ − · − · + + These two relations can be used to eliminate m and obtain a relation in ( , ). h kTo specify the locus conventionally, we use ( , ) xyinstead of ( , ). h kThe equation obtained is (verify) : 2 2 2 2 2 0 3 3 9 a x y ax by + − − + · Example – 11 L OCUS L OCUS L OCUS L OCUS L OCUS 72 Mathematics / Circles Find the equation of the family of circles touching the lines given by 2 2 2 1 0 x y y − + − · Solution: The equation to the pair of lines can be factorised to yield the individual lines as 1 0 x y − + ·and1 0. x y + − · From the following diagram, observe carefully that for a circle to touch both the lines above, its centre must be on one of their angle bisectors because only then will the distance of the centre from the two lines be equal. x + y -=1 0 y x (0,1) ( 1) λ, (0, ) λ x - y + =1 0 For a circle to be able to touch both the lines, its centre must lie on one of their angle bisectors Fig - 58 The angle bisectors can be seen from inspection to be0 x ·and1. y ·Thus, the centre of the variable circle can be assumed to be: (i) (0, ) λ λ ∈! : The distance of (0, ) λfrom the two lines is 1 2 λ − Thus, the equation of the variable circle in this caseis 2 2 2 ( 1) ( ) 2 x y λ λ − + − · ...(1) (ii) ( ,1) λ λ ∈! : The distance of ( ,1) λfrom the two lines in | | 2 λ Example – 12 L OCUS L OCUS L OCUS L OCUS L OCUS 73 Mathematics / Circles Thus, the equation of the variable circle in this case is 2 2 2 ( ) ( 1) 2 x y λ λ − + − · ...(2) The equations (1) and (2) represent the required family of circles. A circle of radius r passes through the origin O and cuts the axes at A and B. Show that the locus of the foot of the perpendicular from O to AB is 2 2 2 2 2 2 1 1 ( ) 4 x y r x y ¸ _ + + · ¸ , Solution: Let the co-ordinates of A and B be ( , 0) aand (0, ) brespectively, so that (like in Example - 8), the equation to the variable circle becomes 2 2 0 x y ax by + − − · A a, ( 0) P h,k ( ) B b (0, ) s O The equation for is + - - = 0 Note that since=, is a diameter of the circle. S x y ax by AOB AB 2 2 Fig - 59 π 2 We have, 2 2 2 4 a b r + · ...(1) Let the foot of perpendicular P have the co-ordinates ( , ). h kSince, OP AB ⊥we obtain 1 k b h a × · − − 2 2 2 2 2 2 2 k h h k h k a b r a b + + ⇒ · · · + 2 2 2 2 2 2 , rk rh a b h k h k ⇒ · · + + ...(2) Using (2) in (1) and ( , ) xyinstead of ( , ), h kwe obtain the required locus. Example – 13 L OCUS L OCUS L OCUS L OCUS L OCUS 74 Mathematics / Circles Find the conditionso that the chordcos sin x y p + · α αsubtends a right angle at the centre of the circle 2 2 2 x y a + · . Solution: x+ y = a 2 22 y x Fig - 60 x + y = p cos sin α α B O A We can obtain the joint equation J to the pair of lines OA and OB by homogenizing the equation of the circle using the equation of the chord AB. 2 2 2 2 cos sin : x y J x y a p ¸ _ + + · ¸ , α α ... (1) For J to represent two perpendicular lines, we must have in (1), 2 2 Coeff. ofoeff. of0 x C y + · which upon simplification yields the required condition as 2 2 2 a p · Suppose that the lines 1 1 1 0 a x b y c + + ·and 2 2 2 0 a x b y c + + ·intersect the co-ordinate axes at points, AB and , C D respectively. Find the condition that must be satisfied if these four points are to be concyclic. Example – 14 Example – 15 L OCUS L OCUS L OCUS L OCUS L OCUS 75 Mathematics / Circles Solution: The co-ordinate of, AB and, C Dcan be evaluated to be 1 1 2 1 1 2 , 0 , 0, , , 0 c c c a b a ¸ _ ¸ _ ¸ _ − − − ¸ , ¸ , ¸ , and 2 2 0, . c b ¸ _ − ¸ , Fig - 61 The four points, and, are given to be concyclic AB CD A C O D B L a x + b y + c = 2 222 :0 L a x + b y + c = 1111 :0 y x Instead of resorting to a detailed calculation, we simply use the result on tangents and secants that we’ve already derived earlier: 2 OAOC ODOB l ⋅ · ⋅ · where l is the length of the tangent drawn from O to the circle. This gives 1 2 2 1 1 2 2 1 c c c c a a b b ¸ _¸ _ ¸ _¸ _ − ⋅ − · − ⋅ − ¸ ,¸ , ¸ ,¸ , 1 2 1 2 0 a a bb ⇒ − · This is the required condition ! L OCUS L OCUS L OCUS L OCUS L OCUS 76 Mathematics / Circles A SSI GN M EN T A SSI GN M EN T A SSI GN M EN T A SSI GN M EN T A SSI GN M EN T [ LEVEL- I ] 1. Find the equation of the circle passing through (–2, 1) and tangent to 3 2 6 0 x y − − ·at (4, 3). 2. If two circles cut a third circle orthogonally, prove that their radical axis will pass through the centre of the third circle. 3. Find the point from which the tangents drawn to the following three circles are of equal lengths : 2 2 1 : 16 16 0 S x y x + − + · 2 2 2 : 3 3 36 8 0 S x y x + − + · 2 2 3 : 16 12 8 0 S x y x y + − − + · 4. Provethatthelengthofthecommonchordofthetwocircles 2 2 2 ( ) ( ) x a y b c − + − · and 2 2 2 ( ) ( ) x b y a c − + − ·is 2 2 4 2( ) c a b − − 5. Provethatthelocusofthecentreofthevariablecirclewhichtouchesexternallythe 2 2 6 6 14 0 x y x y + − − + ·and the y-axis is 2 10 6 14 0. y x y − − + · 6. Findtheequationofthecirclewhoseradiusis3andwhichtouchesinternallythecircle 2 2 4 6 12 0 x y x y + − − − ·at (–1, –1). 7. Provethatthecircle 2 2 2 0 x y ax c + + + · and 2 2 2 0 x y by c + + + · willtouchoneanotherif 2 2 1 1 1 a b c + · . 8. If the circles 2 2 10 16 0 x y x + − + ·and 2 2 2 x y r + ·intersect in real and distinct points, prove that 2 < r < 8. 9. Find the point at which 2 2 4 2 4 0 x y x y + − − − ·and 2 2 12 8 12 0 x y x y + − − − ·touch each other. 10. Find the locus of the mid-point of a variable chord of the circle 2 2 2 2 2 x y x y + − − −which subtends an angle 2 3 π at the centre. 11. Find the locus of the mid point of the variable chord of the circle 2 2 2 x y a + ·which subtends a right angle at(c, 0) 12. If the squares of the lengths of the tangents from a point P to the circles 2 2 2 2 2 2 , x y a x y b + · + ·and 2 2 2 x y c + ·are in A.P., show that a 2 , b 2 and c 2 are in A.P. 13. P is a variable point on y = 4. Tangents from P touch 2 2 4 x y + ·at A and B. The parallelogram PAQB is completed. What is the locus of Q? 14. The angle between T 1 and T 2 is , 3 π where T 1 and T 2 are tangents to 2 2 4 x y + ·and 2 2 9 x y + · respectively. Find the locus of the intersection of T 1 and T 2 . 15. Show that tangents from (–1, 7) to 2 2 25 x y + ·are at right angles to each other. L OCUS L OCUS L OCUS L OCUS L OCUS 77 Mathematics / Circles [ LEVEL- II ] 16. Find the circumcircle of the triangle formed by the three lines 1 : 6 0 L x y + − · 2 : 2 4 0 L x y + − · 3 : 2 5 0 L x y + − · 17. The following two circles intersect in A and B: 2 2 2 1 : 2 0 S x y ax c + + − · 2 2 2 2 : 2 0 S x y bx c + + − · A line through A meets one circle at P while a parallel line through B meets the other circle at Q. Show that the locus of the mid-point of PQ is a circle. 18. A variable circle passes through the fixed point A (a, b) and touches the y-axis. Show that the locus of the other end of the diameter through A is 2 ( ) 4 y b ax − · 19. Suppose that the point P(h, 0) divides a chord AB of the circle 2 2 2 x y a + ·with slopetan , θin the ratio :1. m Prove that 2 2 2 2 cos 1 h h a m m θ − ¸ _ · + ¸ , 20. Find the equation of the circle, the equation of two of whose normal is 2 3 6 2 0 x x y xy + + + ·and which is large enough to just contain the circle 2 2 4 3 . x y x y + · + 21. Find the possible values ofαfor which the point( 1, 1) α − α +lies in the larger segment of the circle 2 2 6 0 x y x y + − − − ·made by the chord2 x y + · 22. Prove that the square of the tangent that can be drawn from any point on one circle to another circle is equal to twice the product of the perpendicular distance of the point from the radical axis of the two circles and the distance between their centres. 23. Find the radius of the smallest circle circumscribing the following three circles : 2 2 1 : 4 5 0 S x y y + − − · 2 2 2 : 12 4 31 0 S x y x y + + + + · 2 2 3 : 6 12 36 0 S x y x y + + + + · 24. A variable chord through a fixed point Q (t, 0 )meets a fixed circle 2 2 2 : , S x y a + · at A and B. Find the locus of a point P on this chord such that QA, QPand QBare in (a) A. P. (b) G.P. 25. If H is the orthocentre of a triangle ABC, prove that the radii of the circumcircles of the trianglesABC and HBC are of the same length. L OCUS L OCUS L OCUS L OCUS L OCUS 78 Mathematics / Circles ANSWERS TRY YOURSELF -I 1. [15,5] 2. ( ) 2 2 2 0 x y mx y c 1 λ + + − · 1 ¸ ] 3. ( ) 2 2 3 3 4 g f c 1 + − 1 ¸ ] 4. 2 2 2 2 8 2 7 0 3 7 12 0 x y x y x y x y 1 + − − + · 1 + − − + · ¸ ] 5. 2 2 1 22 485 17 17 289 x y 1 ¸ _ ¸ _ + + − · 1 ¸ , ¸ , 1 ¸ ] 6. ( ) ( ) 2 2 2 1 1 x y 1 − + + · ¸ ] 7. ( ) 2 2 13 2 13 0 x y y 1 + − − · ¸ ] 8. 2 2 0 x y x y1 + − − · ¸ ] 10. 9 , 2 2 − 1 ¸ _ 1 ¸ , ¸ ] L OCUS L OCUS L OCUS L OCUS L OCUS 79 Mathematics / Circles TRY YOURSELF -II 2. [ ] 11 2 46 0 x y − − · 3. [ ] 4 3 19 0; 4 3 31 0 x y x y + + · + − · 4. [ ] x y a t t · 5. ( ) ( ) 2 2 2 2 2 2 2 0 a b x y x y a b a b 1 ¸ _ + + − + + + · 1 + ¸ , ¸ ] 6. [3x – 2y] =0 7. ( ) ( ) 2 2 6 3 20 x y 1 − + − · ¸ ] 8. [ ] 3 0 x y − · 9. 2 2 4 6 9 0 x y x y1 + + − + · ¸ ] TRY YOURSELF -III 1. 2 2 6 6 9 0 x y x y1 + − − + · ¸ ] 2. [ ] / 4 π 3. 16 31 , 21 63 1 − − 1 ¸ ] 4. 2 2 2 2 2 ax by a b k1 + · + + ¸ ] TRY YOURSELF -IV 1. 2 2 2 2 2 2 18 22 69 0 2 15 0 x y x y x y y 1 + − − + · 1 + − − · ¸ ] 2. ( )( ) ( ) 2 2 2 2 2 x y a b abbx ay 1 + + · + ¸ ] 3. ( ) 2 2 3 23 4 35 0 x y x y 1 + − · + · ¸ ] 4. 2 2 3 2 0 x y x y1 + + + · ¸ ] 5. 2 2 7 11 38 0 x y x y1 + + − + · ¸ ] L OCUS L OCUS L OCUS L OCUS L OCUS 80 Mathematics / Circles ASSIGNMENT ANSWER LEVEL - I 1. ( ) 2 2 7 4 82 55 0 x y x y 1 + + − + · ¸ ] 3. 10 2 , 3 3 1 ¸ _ − 1 ¸ , ¸ ] 6. 2 2 5 5 8 14 32 0 x y x y1 + − − − · ¸ ] 9. 2 4 , 5 5 1 ¸ _ − − 1 ¸ , ¸ ] 10. 2 2 2 2 1 0 x y x y1 + − − + · ¸ ] 11. ( ) 2 2 2 2 2 2 0 x y cx c a 1 + − + − · ¸ ] 13. ( )( ) 2 2 2 2y x y x y 1 · + + ¸ ] 14. ( ) 2 2 3 76 x y 1 + · ¸ ] LEVEL II 16. 2 2 17 19 50 0 x y x y1 + − − + · ¸ ] 17. ( ) 2 2 0 x y a b x1 + + + · ¸ ] 20. 2 2 6 3 45 0 x y x y1 + + − − · ¸ ] 21. ( ) 1,1 −1 ¸ ] 23. [approx7.3] 24. ( ) ( ) 2 2 2 2 2 2 0 a x y tx b x y tx a 1 + · 1 + − + · 1 ¸ ] L OCUS L OCUS L OCUS L OCUS L OCUS 81 Mathematics / Circles