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https://docs.google.com/document/d/15rALinbLkiV3640LZPeOwn2Y_jpibHw7/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Circle Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation of a circle in the parametric form, equation of a circle when the end points of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to the circle, length of the tangent, equation of the tangent, equation of a family of circles through the intersection of two circles, condition for two intersecting circles to be othogonal. Among all curves, circle is the simplest curve. It is common in almost every sphere of life. A wheel, a circular object, has revolutionised the transportation like all motor vehicles, railways. pullies, gears, various rings are all circular. CIRCLE STANDARD EQUATION A circle is the locus of a point which moves such that its distance from a fixed point, called as centre, is equal to a given distance, called as radius. Thus, equation to a circle with centre (α, β) and radius a is : (x − α)2 + (y − β)2 = a2 If the centre is origin then x2 + y2 = a2 Give the equation of shaded region in rectangular co-ordinates. Solution : The centre of both circles is (2, 3) and radii are 4 and 6. Thus the equation for the shaded region between two circles shall be 4 ≤ OP ≤ 6 ⇒ General Equation of Circle The general equation of a circle is : S ≡ x 2 + y 2 + 2gx + 2fy + c = 0 C H A P T E R CHAPTER INCLUDES : Circle and its standard equation General form of equation of circle Different forms of equation of circle Intercepts of circle Position of a point with respect to circle Tangents of a circle and its length Angle of intersection of two circles Equation of a circle through intersection of a line and a circle Equation of a circle through intersection of two circles Common tangents to two circles Radical axis, chord of contact and chord with middle point It has three arbitrary constants. Its centre is (–g, –f) and radius is Solved examples . The circle is real, point circle or imaginary according as g 2 + f 2 − c > 0, = 0, or < 0; c = 0 if circle passes through origin. Give the equation of circle whose ordinate of centre is double of its abscissa and radius is equal to the sum of both. Solution : Let the centre be (–g , –f,) and radius is . So according to given condition f = 2g and = 3g or g2 + f 2 – c = 9g2 ∴ c = g2 + 4g2 – 9g2 = –4g2 ∴ Equation of required circle is x2 + y2 + 2gx + 2fy + c = 0 ⇒ x2 + y2 + 2gx + 4gy – 4g2 = 0 or (x + g)2 + (y + 2g)2 = 9g2 General Equation of Second Degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 shall represent a circle if Coefficient of x2 & y2 are equal i.e. a = b ⎤  & Coefficient of xy is zero i.e. h = 0 ⎦ DIFFERENT FORMS OF EQUATIONS OF CIRCLES Parametric Form Circle with centre (α, β) and radius a is (x – α)2 + (y – β)2 = a2 It can be represented in parametric form as x = α + acosθ y = β + asinθ ⎤ θ is parameter ⎥ 0 ≤ θ ≤ 2π A circle with centre (0, 0) and radius a is x = acosθ y = asinθ ⎤ θ is parameter  Represent the circle x2 + y2 + 4x – 6y – 3 = 0 in parametric form. Solution : x2 + y2 + 4x – 6y – 3 = 0 (x + 2)2 + (y – 3)2 = 16 Centre (–2, 3); radius (a) = 4 ∴ x = –2 + 4 cosθ y = 3 + 4 sinθ ⎤ is the parametric form  Parametrically a circle is x = –1 + 3 sinθ and y = – 4 + 3 cosθ. Give the cartesian equation of circle and find its area. Solution : sin θ = x + 1; cos θ = y + 4 3 3 ⇒ sin2 θ + cos2 θ = 1 (x + 1)2 + (y + 4)2 = ∴ 9 9 1 ⇒ (x + 1)2 + (y + 4)2 = 9 Centre (–1, –4) and radius = 3 Hence area is πr2 = 9π sq. units Find the shortest and longest distance between following two circles x = cosθ; y = sinθ and x = 3 + 2cosθ; y = 3 + 2sinθ Solution : Shortest distance d between any two circles is c1c2 – r1 – r2 and longest distance is c1c2 + r1 + r2 c1 = (0, 0); c2 = (3, 3), r1 = 1, r2 = 2 ∴ c1c2 = = 3 ∴ Shortest distance = 3 − 1− 2 = 3( − 1) Longest distance = 3( + 1) Equation of Circle in Diametric Form If two diametrically opposite points on a circle are (x1, y1) ⎛ x + x y + y ⎞ (x2,y2) (x1,y1) and (x , y ) then centre = ⎜ 1 2 , 1 2 ⎟ 2 2 ⎝ 2 2 ⎠ Radius = and equation of circle is (x–x1) (x–x2) + (y–y1) (y–y2) = 0 A circle with centre origin and a point on its periphery is (3, 0). Find the equation of circle in diametric form. Solution : Since centre is (0, 0) and point A is (3, 0). Hence diametrically opposite point B will be (–3, 0) ∴ (x – 3) (x + 3) + (y – 0) (y – 0) = 0 or (x – 3) (x + 3) + y2 = 0 Equation of circle with centre (α, β) and touching the x–axis As it is clear from figure that r = β ∴ (x – α)2 + (y – β)2 = β2 or x2 + y2 – 2αx – 2βy + α2 = 0 Equation of circle with centre (α, β) and touching the y–axis Here r = α ∴ (x –α)2 + (y –β)2 = α2 ⇒ x2 + y2 – 2αx – 2βy + β2 B A (3, 0) Equation of circle with radius a and touching both the axes There will be four such circles There centres are (±a, ±a) and radius a. ∴ (x ± a)2 + (y ± a)2 = a2 ⇒ x2 + y2 ± 2ax ± 2ay + a2 = 0 (–a, a) (–a, a) (a, a) (a, –a) Find the equation of a circle in third quadrant of radius 3 touching the y-axis at (0, –4). Solution : As shown in figure, the required circle will have its centre as Hence (x + 3)2 + (y + 4)2 = 9 ∴ x2 + y2 + 6x + 8y + 16 = 0 ( f ) Circle through three non-collinear points (x1, y1), (x2, y2) and (x3, y3) Let the circle be x2 + y2 + 2gx + 2fy + c = 0 Then all three points must satisfy it x 2 + y 2 + 2gx + 2fy + c = 0 ⎤ 1 1 1 1 ⎥ ⇒ x 2 + y 2 + 2gx2 + 2y 2 + c = 0 ⎥ simultaneous solution of these equations x 2 + y 2 + 2gx + 2fy + c = 0⎥ 3 3 3 3 ⎦ shall give g, f & c and hence we can write the equation of circle. Alternate : The equation of circle is given by 1 1 1 1 = 0 2 2 2 2 3 3 3 3 Give the equation of circle passing through (3, 0), (0, 3) and origin (0, 0). x 2 + y 2 x y 1 x 2 + y 2 x y 1 x 2 + y 2 x y 1 x 2 + y 2 x y 1 Illustration 8 : Solution : x 2 + y 2 9 9 0 x y 1 3 0 1 = 0 0 3 1 0 0 1 x 2 + y 2 9 ⇒ 0 0 x y 3 0 − 3 3 0 0 1 1 = 0 0 1 R3 − R2 x 2 + y 2 9 ⇒ 0 0 x y 3 0 − 3 3 0 0 1 0 = 0 0 1 R2 − R4 x 2 + y 2 ⇒ 9 0 x y 3 0 = 0 − 3 3 ⇒ (x2 + y2) (9) – x (27) + y (–27) = 0 Intercepts Made by a Circle on Axes Intercept made on x-axis by S ≡ 0 is : 2 Intercept made on y-axis by S ≡ 0 is : 2 If g2 = c (resp., f 2 = c), the circle touches the x-axis (resp. y-axis). if c = g2 = f 2 then circle touches both axes. Length of the chord intercepted by Intercepts are always positive. x 2 + y 2 = a2 on the line y = mx + c is 2 Give the equation of a circle passing through origin and having intercept on x-axis as double of intercept on y-axis and radius 4 units. Solution : As the circle x2 + y2 + 2gx + 2fy + c = 0 passes through origin hence c = 0. ∴ x intercept = 2 | g | y intercept = 2 | f | ∴ 2 | g | = 2 × 2 ∴ | g | = 2 | f | i.e. g = ±2f Radius = 4 = ⇒ 16 = 5f 2 f = ± = ∴ g = ± Hence circle is x 2 + y 2 ± x ± y = 0 i.e. four such circles should exist. Position of a Point with respect to Circle A point P(x1, y1) lies inside, on or outside the circle S ≡ x 2 + y 2 + 2gx + 2fy + c = 0 according as x 2 + y 2 + 2gx + 2fy + c is < 0, = 0 or > 0. 1 1 1 1 TANGENT TO A CIRCLE Corresponding to any point P(x1, y1), we define the expressions S1 ≡ x 2 + y 2 + 2gx 2fy + c T ≡ xx1 + yy1 + g(x + x1 ) + f (y + y1) + c Equation of tangent at (x1, y1) to S ≡ 0 is : T = 0 Equation of normal at (x1, y1) to S ≡ 0 is y (x1 + g ) − x (y1 + f ) + fx1 − gy1 = 0 The line y = mx + c intersects the circle real and distinct points if c2 < a2(1 + m2) x 2 + y 2 = a2 at real and coincident points if c2 = a2(1 + m2) imaginary points if c2 > a2(1 + m2) ∴ y = mx + c is a tangent to x 2 + y 2 = a2 if c 2 = a2(1+ m2 ) ⎛ − am a ⎞ The line y = mx + a is a tangent to x 2 + y 2 = a2 at ⎜ , ⎟ ⎜ 1+ m 2 1+ m 2 ⎟ ⎛ am − a ⎞ The line y = mx − a is tangent to x 2 + y 2 = a2 at ⎜ , ⎟ . Length of the tangent from P(x1, y1) to S ≡ 0 is equal to ⎜ ⎝ S1 , 1+ m 2 1+ m 2 ⎟ where P lies outside S ≡ 0. Equation to the pair of tangents from P(x1, y1) to S ≡ 0 is given by SS1 = T 2 . With respect to circle x2 + y2 + 4x – 6y + 8 = 0, find the tangent passing through point (4, –2). Solution : The point (4, –2) is lying outside the circle as 16 + 4 + 16 + 12 + 8 > 0 S1 > 0 A line passing through (4, –2) is y + 2 = m (x–4) or y = mx – 4m – 2 to be a tangent the distance of line from centre (–2, 3) should be equal to radius = ∴ −2m − 4m − 2 − 3 = ± ⇒ (6m + 5)2 = 5 + 5m2 ⇒ 36m2 + 25 + 60m = 5 + 5m2 ⇒ 31m2 + 60m + 20 = 0 ⇒ m = − 60 ± 3600 − 2480 62 = − 60 ± 1120 62 = − 30 ± 31 280 = − 30 ± 2 70 31 ⎛ − 30 ± 2 70 ⎞ ∴ y = ⎜ ⎟(x − 4) + 2 is the required equations of tangents. ⎜ 31 ⎟ Angle of Intersection of Two Circles Angle of intersection of two circles S1 = 0 and S2 = 0 Let and S1 ≡ x 2 + y 2 + 2g1x + 2f1y + C1 S2 ≡ x 2 + y 2 + 2g2 x + 2f2 y + C2 Let S1 and S2 have centres C1 and C2 respectively and radii r1 and r2 respectively. The angle of intersection θ of the circles is the angle between their tangents at points of intersection and is given by r 2 + r 2 − (C C )2 2g1g2 + 2f1f2 − C1 − C2 cos θ = 1 2 1 2 2r1r2 2 Orthogonal circles : Circles S1 ≡ 0 and S2 ≡ 0 are said to intersect orthogonally if θ = 90° ; i.e., if 2g1g2 + 2f1f2 = C1 + C2 Find the equation of circle passing through origin and orthogonally intersecting the circles. x2 + y2 + 2x + 4y + 2 = 0 and x2 + y2 + 4x + 6y – 3 = 0 Solution : Let the circle be x2 + y2 + 2gx + 2fy = 0 Since it intersects the above two circles orthogonally, hence 2.g.1 + 2.f.2 = 0 + 2 or and 2g.2 + 2 f. 3 = 0 – 3 ... (i) ... (ii) e.q., (i) & (ii) when solved simultaneously give g = –6 and f = 7 2 ∴ The required circle is x2 + y2 – 12x + 7y = 0 Equation of a Circle through Intersection Points of Circle and Line Let the circle be S = 0 and line be L = 0 S + λL = 0 is a circle which passes through the points of intersection of both. Equation of any circle that passes through two given points (x1, y1) and (x2, y2) is x (x − x1)(x − x2 ) + (y − y1)(y − y 2 ) + λ x1 x2 Let u ≡ ax + by + k = 0 y 1 y1 1 = 0 y 2 1 If u ≡ 0 is a tangent to S ≡ 0 at the point P then S + λu ≡ 0 is the equation of circles touching S ≡ 0 at P. Equation of the circles which touch the line u ≡ 0 at (x1, y1) is (x − x1)2 + (y − y1)2 + λu ≡ 0 Equation of a Circle through Intersection of Two Circles If S1 ≡ 0 and S2 ≡ 0 be two circle, intersecting in real points, then S1 + λS2 = 0 ( λ ≠ −1) is the equation of the family of circles passing through the common points of S1 ≡ 0 and S2 ≡ 0 . If S1 ≡ 0 and S2 ≡ 0 intersect, then S1 − S2 ≡ 0 is the equation of their common chord. If S1 ≡ 0 and S2 ≡ 0 touch each other then S1 − S2 ≡ 0 is the equation of their common tangent at the point of contact. Common Tangents to Two Circles Let circles S1 ≡ 0 and S2 ≡ 0 have centers O1 and O2 respectively and radii r1 and r2 respectively. The number of common tangents of S1 ≡ 0 and S2 ≡ 0 is zero, if one circle lies totally inside the other O1O2 < r1 − r2 one, if circles touch each other internally O1O2 = | r1 − r2 | two, if circles intersect at two distinct points | r1 − r2 | < O1O2 < r1 + r2 three, if circles touch each other externally O1O2 = r1 + r2 four, if circles lie outside each other Two of the tangents are direct common tangents. Two of the tangents are transverse common tangents. T O O > r + r 1 1 2 1 2 If direct common tangents meet the line O1O2 in T1 then T1 divides the segment O1O2 externally in the ratio r1 : r2. If the transverse common tangents meet the line O1O2 in T2 then T2 divides the segment O1O2 internally in the ratio r1 : r2. The coordinates of T1 and T2 having been found, the corresponding tangents are straight lines through it such that perpendiculars on them from O1 are each equal to r1. Radical Axis, Chord of Contact and Chord with the given Middle Point The radical axis of S1 ≡ 0 and S2 ≡ 0 is the locus of a point which moves such that the lengths of the tangents drawn from it to the two circles are equal. The equation to the radical axes is : S1 – S2 = 0. Equation of the pair of tangents : Combined equation of the tangents PQ and PR drawn from P(x1, y1) to the circle S = 0 is given by S = x 2 + y 2 + 2gx + 2fy + c , where S1 = x 2 + y 2 + 2gx + 2fy + c 1 1 1 1 T = xx1 + yy1 + g(x + x1 ) + f (y + y1) + c Equation of the chord whose middle point is given : If P(x1, y1) be the middle point of a chord AB of the circle S = 0, then equation of the chord is given by i.e., xx1 + yy1 + g(x + x1) Chord of contact It is the line joining the points of contact of the tangents drawn from a point outside the circle. If from a point P(x , y ), two tangents PA and PB are (x1, y1) 1 1 drawn. The line AB is the chord of contact of the point P with respect to the given circle. Equation of the chord of contact with respect to the Circles x2 + y2 = a2 is xx1 + yy1 = a2 Circle x2 + y2 + 2gx + 2fy + c = 0 is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 Length of the tangent from a point to the circle T Let P(x1, y1) be a point and PT is the tangent drawn from P to the circle S ≡ x2 + y2 – a2 = 0, then a P O PT = (x1, y1) (o, o) = If the circle is S ≡ x2 + y2 + 2gx + 2fy + c = 0, then PT = = Image of the circle through a line Let S ≡ (x – h)2 + (y – k)2 – a2 = 0 Radius of the image circle will be same as the radius of S, but centre will be changed Centre of the given circle (h, k) Centre of the image circle through y-axis is c1(–h, k) Centre of the image circle through x-axis is c2(h, –k) y (h′, k′) B C′ For the image through the line AB. Let equation of AB is x + my + n = 0 and let centre of the image circle be ( h′, k′). Then x (–h, k) C1 A C (h, k) x′ m of cc′ × m of AB = –1 ⇒ ⎛ k – k′ ⎞ × ⎛– λ ⎞ = –1 C ⎜ ⎟ ⎜ ⎟ 2 ⎝ h – h′ ⎠ ⎝ m ⎠ (h, –k) cc′ = ⎛ h + h′ , k + k′ ⎞ and middle point of ⎜ ⎝ 2 2 ⎟ lies on y′ ⎠ x + my + n = 0 solving for h′ and k′. Example 1 : Find the equation of the circle circumscribing the triangle formed by the lines, x + y = 6, 2x + y = 4 and x + 2y = 5. Solution : x + y = 6 …(1) 2x + y = 4 …(2) x + 2y = 5 …(3) The vertices of the triangle ABC formed by (1), (2) and (3) are : A(–2, 8) ; B(1, 2) ; C(7, –1) Let x 2 + y 2 + 2gx + 2fy + c = 0 be the equation of the required circle ; then, since, A, B and C lie on this circle, we get : –4g + 16f + c = –68 …(4) 2g + 4f + c = –5 …(5) 14g – 2f + c = –50 …(6) By (4) and (5) we get : 2g – 4f = 21 …(7) By (5) and (6) we get : –4g + 2f = 15 …(8) Solving (7) and (8), we have : g = − 17 2 and f = − 19 2 ∴ By (5), c = 50 and thus, x 2 + y 2 − 17x − 19y + 50 = 0 is the required equation. Example 2 : Find the equations of circles which have radius and which touch the line 2x – 3y + 1 = 0 at (1, 1). Solution : Let one of the circles have centre C1(x1, y1) and let the point P have coordinates (1, 1). Since C1P ⊥ 2x – 3y + 1 = 0. The equation of C1P is 3x + 2y = 5 Since C1(x1, y1) lies on it, we have : 3x1 + 2y1 = 5 …(1) Also, C1P = 13 ⇒ = ∴ 2x1 − 3y1 + 1 = ± 13 Taking ‘+’ sign : 2x1 – 3y1 = 12 …(2) Taking ‘–’ sign : 2x1 – 3y1 = –14 …(3) Solving (1) and (2) we get : x1 = 3 and y1 = –2 ∴ Equation of one of the circles is (x − 3)2 + (y + 2)2 = 13 Solving (1) and (3), we get : x1 = –1 ; y1 = 4 ∴ Equation of the other circle is (x + 1)2 + (y − 4)2 = 13 Example 3 : If the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 cut the coordinate axes in concyclic points, show that a1a2 = b1b2. Solution : Let the straight line a1x + b1y + c1 = 0 y cut the coordinate axes in the points A and B respectively ; then, A and B ⎛ − c1 ⎞ ⎛ − c1 ⎞ have coordinates ⎜ ⎝ , 0⎟ a1 ⎠ and ⎜ 0, ⎝ ⎟ b1 ⎠ respectively. x Let the line a2 x + b2 y + c2 = 0 cut the axes in points C and D respectively ; then, C and D have coordinates ⎛ − c2 ⎞ ⎛ − c2 ⎞ ⎜ a , 0 ⎟ and ⎜ 0, b ⎟ respectively. ⎝ 2 ⎠ ⎝ 2 ⎠ By geometry, since A, B, C and D are concyclic, we have l (OA) ⋅ l (OC) = l (OB) ⋅ l (OD) ⇒ ⋅ = ⋅ ⇒ a1a2 = b1b2 Example 4 : Find the equation of the circles which pass through the origin and cut off chords of length ‘a’ from each of the lines y = x and y = –x. Solution : Let x 2 + y 2 + 2gx + 2fy + c = 0 …(1) be the equation of such circle. We note that ‘c = 0’ as (1) passes through the origin. Let OP and OQ denote the chords intercepted by (1) on y = x and y = –x respectively where O is the origin. If the coordinates of P are (α, α) then “a = OP ” ⇒ The possible coordinates of Q are (–α, α) or (α, –α) α = ± a Since P and Q lie on (1), we get : g = 0 and f = –α or g = –α and f = 0 according as Q is (–α, α) or (α, –α) and where P is (α, α) ∴ Equation of the required circles are : when α = a : x 2 + y 2 − 2 ay = 0 or x 2 + y 2 − 2 ax = 0 when α = − a : x 2 + y 2 + 2ay = 0 or x 2 + y 2 + 2a ⋅ x = 0 Example 5 : A triangle has two of its sides along the coordinate axes ; its third side touches the circle x 2 + y 2 − 2ax − 2ay + a2 = 0 . Prove that the locus of the circumcentre of the triangle is Solution : a2 − 2a (x + y ) + 2xy = 0 where a > 0 x 2 + y 2 − 2ax − 2ay + a2 = 0 Eq. (1) has center C(a, a) and radius r = a. …(1) Let OAB be the required triangle and let M(x1, y1) be any point on the locus ; then, M is the mid point of segment AB. A ≡ (2x1, 0) ; B ≡ (0, 2y1 ) Equation of the straight line AB is i.e., xy1 + yx1 − 2x1y1 = 0 x 2x1 + y = 1 2y1 Since AB is a tangent to the circle, the length of the perpendicular from C on AB is equal to radius of the circle ∴ = a ∴ (ay1 + ax1 − 2x1y1)2 = a2(x 2 + y 2 ) 1 1 4x 2y 2 + 2a2x y − 4ax y (x + y ) = 0 1 1 1 1 1 1 1 1 ∴ 2x1y1 + a2 − 2a (x1 + y1) = 0 (Θ x1 ≠ 0 ; y1 ≠ 0) x Equation of locus is a2 − 2a(x + y ) + 2xy = 0 Example 6 : Through a fixed point (h, k) secants are drawn to the circle x 2 + y 2 = r 2. Show that locus of mid points of portions of secants intercepted by the circle is x 2 + y 2 = hx + ky . Solution : x 2 + y 2 = r 2 …(1) B P(h, k) Eq. (1) has centre C(0, 0) and radius r. A M Let P ≡(h, k ) and let M ≡ (x1, y1) be any point on the C(0, 0) locus ; then, M is the mid point of the chord AB which is part of the secant drawn from P(h, k) to (1). Since CM ⊥ AB, we have : y1 − 0 ⋅ y1 − k = −1 x1 − 0 ∴ y1(y1 − k ) + x1(x1 − h) = 0 x1 − h x 2 + y 2 = hx + ky 1 1 1 1 Equation of locus is x 2 + y 2 = hx + ky Example 7 : Two circles, each of radius 5 units, touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, find the equation of the circles. Solution : Let C1 and C2 denote the centres of the two circles. The equation of the common tangent at P(1, 2) is 4x + 3y = 10 …(1) ∴ The equation of the common normal at (1, 2) is 3x – 4y + 5 = 0 …(2) tan θ = slope of (2) = 3 ; 4 ∴ cos θ = 4 5 and sin θ = 3 5 Also, equation of normal in parametric form is x − 1 = r cos θ; y − 2 = r sin θ …(3) Since PC1 = PC2 = 5 units, the coordinates of C1 and C2 are obtained by putting r = 5 and r = –5 successively in (3). Hence, C1(5, 5) and C2(–3, –1) are the coordinates of the centres. The equation of the two circles are: (x − 5)2 + (y − 5)2 = 25 and (x + 3)2 + (y + 1)2 = 25 at a point on the circle x 2 + y 2 = a2 intersects a concentric circle C at two points P and Q. The tangents to the circle C at P and Q meet at a point on the circle circle C. x 2 + y 2 = b2 . Find the equation to the Solution : Let the equation of circle C be x 2 + y 2 = r 2 …(1) Let A(h, k) be a point on the circle x 2 + y 2 = a2 …(2) Tangent at A to (2) cuts the circle (1) at P and Q. Let the tangents at P and Q meet at B(b cos θ, b sin θ) on the circle x 2 + y 2 = b2 Equation of chord of contact PQ of tangents drawn from B to (1) is b x cos θ + b y sin θ = r 2 Since (4) is a tangent at A to (2) …(3) …(4) ∴ a = r ∴ r2 = ab and eq. (1) becomes Example 9 : Consider a family of circles passing through two fixed points A(3, 7) and B(6, 5). Show that chords in which the circle x 2 + y 2 − 4x − 6y − 3 = 0 cuts the members of the family are concurrent at a point. Find the coordinates of this point. Solution : Equation of the chord AB is 2x + 3y = 27 …(1) Equation of family of circles through A and B in (x − 3) (x − 6) + (y − 7)(y − 5) + λ (2x + 3y − 27) = 0 x 2 + y 2 − 9x − 12y + 53 + λ (2x + 3y − 27) = 0 …(2) x 2 + y 2 − 4x − 6y − 3 = 0 …(3) The common chords of (2) and (3) are given by : 5x + 6y − 56 − λ (2x + 3y − 27) = 0 which is a family of lines through the points of intersection of 5x + 6y – 56 = 0 and 2x + 3y – 27 = 0. ⎛ Their point of intersection is ⎜ ⎝ 23 ⎞ ⎟ . ⎠ Example 10 : Find the equation of the circle through the points of intersection of the circles x 2 + y 2 − 4x − 6y − 12 = 0 and x 2 + y 2 + 6x + 4y − 12 = 0 and intersecting the circle x 2 + y 2 − 2x − 4 = 0 orthogonally. Solution : Equation of the family of circles through the points of intersection of the circles x 2 + y 2 − 4x − 6y − 12 = 0 and x 2 + y 2 + 6x + 4y − 12 = 0 is given by : x 2 + y 2 − 4x − 6y − 12 + λ (x2 + y 2 + 6x + 4y −12) = 0 i.e. x 2 + y 2 + (6λ − 4) x + (4λ − 6) y − 12 = 0 …(1) 1+ λ 1+ λ Since one of the circle given by (1) intersects the circle orthogonally, we have, (for that value of λ), 2 ⋅ ⎛ 3λ − 2 ⎞ x 2 + y 2 − 2x = 4 ⎜ 1+ λ ⎟(−1) + 0 ⎠ = –12 – 4 3λ – 2 = 8 (1 + λ) 5λ = – 10 ; λ = –2 Equation of the required circle is : x 2 + y 2 + 16x + 14y − 12 = 0 ❑ ❑ ❑