Circle Solved Examples

Circle (ii) In the above equation DEFINITION Circle is locus of a point which moves at a constant distance from a fixed point . This constant distance is called radius of the circle and fixed point is called centre of the circle. STANDARD FORMS OF EQUATION OF A CIRCLE General Form  If c = 0 the circles passes through the origin  If f = 0 the centre is on x-axis  If g = 0 the centre is on y-axis (iii) The general eqaution of second degree ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a circle if a = b  0 and h = 0. Note : The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0, where g,f,c are constants. That every second degree equation in x & y, in which coefficient of x2 is equal to coefficient of y2 & the coefficient of xy is zero, always represents a circle. For this circle, Centre = ( – g , –f) = , Radius = Ex.1 Find the radius of the circle x2 + y2 + 4x – 6y + 1 = 0. Sol. Here Note : (i) The above equation represents x2 + y2 + 4x – 6y + 1 = 0 on comparing from general eqn.  a real circle if g2 + f2 > c 2g = 4  a point circle if g2 + f2 = c 2f = –6  f = –3,  an imaginary circle if g2 + f2 < c   g = 2, c=1 r = g2  f 2  c  4  9  1  12  2 3 49 Circle Ex.2 If (4, –2) is the one extremity of diameter to the Note : The equation of a circle with radius 'a' touching circle x2 + y2 – 4x + 8y – 4 = 0 then find its other x-axis at the origin is extremity. x2 + (y ± a)2 = a2     x2 + y2 ± 2ay = 0 Sol. Centre of circle is (2, –4). Let the other extremity is (h, k) (e) Which touches y-axis : The equation of a circle with radius 'a' touching y-axis at a distance k from  (h, k) = (0 , –6) the origin is (x – a)2 + (y – k)2 = a2 Ex.3 If y = 2x + m is a diameter to the circle x2 + y2 + 3x + 4y – 1 = 0, then find m Sol. Centre of circle = ( –3/2 , –2). This lies on diameter y = 2x + m – 2 = –3/2 × 2 + m  m=1 EQUATION OF A CIRCLE IN SOME Note : The equation of a circle with radius 'a' touching SPECIAL CASES y-axis at the origin is (a) The circle with centre as origin & radius ‘r’ has the (x ± a)2 + y2 = a2  x2 + y2 ± 2ax = 0 2 2 2 equation; x + y = r . Ex.4 If the circle x2 + y2 + 2gx + 2fy + c = 0 touches x-axis, then find the value of c. Sol. Touches x-axis, hence radius = ordinate of centre. (b) The circle with centre (h, k) & radius ‘r’ has the equation; (x  h)2 + (y  k)2 = r2. Hence 2 g2  f 2  c  (  f ) or g = c. Ex.5 Find the equation of a circle whose centre is (3, –1) and radius is 3 Sol. (x – 3)2 + (y + 1)2 = 32 x2 – 6x + 9 + y2 + 2y + 1 = 9 x2 + y2 – 6x + 2y + 1 = 0 (c) Which touches both axes : The equation of a circle Ex.6 Find the equation of a circle with centre at the origin with radius 'a' touching both coordinate axes is and which passes through the point ( , ) . given by (x ± a)2 + (y ± a)2 = a2 Sol. Here radius = ; so the required equation is 2 2 2 2 x +y = + (d) Which touches x-axis : The equation of a circle with radius 'a' touching x-axis at a distance h from the origin is (x – h)2 + (y – a)2 = a2 The equation of circle with (x1, y1) & (x2, y2) as extremeties of its diameter is: (x  x1) (x  x2) + (y  y1) (y  y2) = 0. 50 Circle This is obtained by the fact that angle in a semicircle Ex.9 Find the equation of the circle, the coordinates of is a right angle. the end points of whose diameter are (–1, 2) and (4, –3)  (Slope of PA) (Slope of PB) = – 1 y  y1 y  y2 Sol. We know that the equation of the circle described  x  x1 . x  x 2 = – 1 on the line segment joining (x1, y1) and (x2, y2) as a  (x – x1) (x – x2) + (y – y1) (y – y2) = 0 diameter is (x – x1) (x – x2) + (y – y1) (y – y2) = 0. Note that this will be the circle of least radius Here, x1 = –1, x2 = 4, y1 = 2 and y2 = –3. passing through (x1, y1) & (x2, y2). So, the equation of the required circle is (x + 1) (x – 4) + (y – 2) (y + 3) = 0  x2 + y2 – 3x + y – 10 = 0. Ex.7 Find the equation of the circle whose centre is Parametric Equation of a Circle (1, –2) and radius is 4. (a) The parametric equations of a circle Sol. The equation of the circle is 2 2 2 (x – 1) + (y – (–2)) = 4 x2 + y2 = a2 are x = acos , y = asin.  (x – 1)2 + (y + 2)2 = 16 Hence parametric coordinates of any point lying on 2 2 the circle x2 + y2 = a2 are (acos, asin)  x + y – 2x + 4y – 11 = 0 Ex.8 Find the equation of the circle which passes through the point of intersection of the lines 3x – 2y – 1 = 0 and 4x + y – 27 = 0 and whose centre is (2, – 3). Sol. Let P be the point of intersection of the lines AB and LM whose equations are respectively 3x – 2y – 1 = 0 ..........(i) and 4x + y – 27 = 0 ..........(ii) (b) The parametric equations of the circle Solving (i) and (ii), we get x = 5, y = 7. (x–h)2 + (y – k)2 = a2 are So, coordinates of P are (5, 7). Let C(2, –3) be the x = h + a cos , y = k + asin. centre of the circle. Since the circle passes through Hence parametric coordinates of any point lying on P, therefore the circle are (h + acos, k + asin) CP = radius  (5  2)2  (7  3 )2 (c) Parametric equations of the circle  radius = 109 . x2 + y2 + 2gx + 2fy + c = 0 is Hence the equation of the required circle is x=–g+ cos, (x – 2)2 + (y + 3)2 = 109 2 y=–f+ sin Ex.9 Find the centre & radius of the circle whose equation is x2 + y2 – 4x + 6y + 12 = 0 Sol. Comparing it with the general equation x2 + y2 + 2gx + 2fy + c = 0, we have 2g = – 4  g = –2 2f = 6  f=3 & c = 12  centre is (–g, –f) i.e. (2, –3) and radius = g2  f 2  c = ( 2)2  (3)2  12 = 1 Ex.10 Find the parametric coordinates of any point of the circle x2 + y2 + 2x – 3y – 4 = 0 Sol. Centre = radius = Parametric coordinates of any point are (–1 + 7 3 cos , 7 3 + sin) 3 2 51 Circle Ex.11 Find the parametric equations of the circle x2 + y2 – 4x – 2y + 1 = 0 Sol. We have : x2 + y2 – 4x – 2y + 1 = 0  (x2 – 4x ) + (y2 – 2y) = – 1  (x – 2)2 + (y – 1)2 = 22 So, the parametric equations of this circle are x = 2 + 2 cos  , y = 1 + 2 sin . CONDITION OF TANGENCY A line L = 0 touches the circle S = 0, if length of perpendicular drawn from the centre of the circle to the line is equal to radius of the circle i.e., p = r. This is the condition of tangency for the line L = 0. The line y = mx + c touches the circle x 2 + y2 = a2 if c = ± a 1 m2 Ex.12 Find the equations of the following curves in cartesian form. Also, find the centre and radius of the circle x = a + c cos , y = b + c sin  Sol. We have : x = a + c cos , y = b + c sin x a c , sin  = cos  =  xa  y b   +   = cos2 + sin2 c    c  2 the line y = mx ± a 1 m 2 is a tangent of the circle x2 + y2 = a2 and its point  2 2 Note : 2 (x – a) + (y – b) = c  Clearly, it is a circle with centre at (a, b) and radius c.  POSITION OF A POINT AND LINE WITH RESPECT TO A CIRCLE 1. Position of a point A point (x1, y1) lies outside, on or inside a circle S  x2 + y2 + 2gx + 2fy + c = 0 according as S1  x12 + y12 + 2gx1 + 2fy1 + c is positive, zero or negative. So  S1 > 0  (x1, y1) is outside the circle  S1 = 0  (x1, y1) is on the circle  S1 < 0  (x1, y1) is inside the circle 2. Position of a line Let L = 0 be a line and S = 0 be a circle. If 'a' be the radius of the circle and 'p' be the length of the perpendicular from its centre on the line, then  p > a  line is outside the circle  p = a  line touches the circle  p < a  line is a chord of the circle  p = 0  line is a diameter of the circle    am ,  a  of contact is   2 1  m2   1 m y b c  2 Thus, for every value of m, If a2 (1 + m2) – c2 > 0 line will meet the circle at real and different points.  If c2 = a2 (1 + m2) line will touch the circle.  If a2 (1 + m2) – c2 > 0 line will meet circle at two imaginary points. Ex13 For what value of c will the line y = 2x + c be a tangent to the circle x2 + y2 = 5 ? Sol. We have : y = 2x + c or 2x – y + c = 0 ......(i) and x2 + y2 = 5 ........(ii) If the line (i) touches the circle (ii), then length of the  from the centre (0, 0) = radius of circle (ii) 20 0  c  = 5  =± 5  2  (1) c  2 2 5 c 5 = 5 c=±5 Hence, the line (i) touches the circle (ii) for c = ± 5 Ex.14 If the line y = mx + c touches the circle x2 + y2 = 4y then find c. Sol. Centre of the circle = (0, 2), radius = 2. So condition of tangency c2 p = a  m2  1 2  c=2 (1  m2  1) 52 Ex.15 Find the point at line y = x + 2a touches the circle x2 + y2 = a2 . Sol. The line y = mx + a 1  m2 touches the circle x2 + y2 = a2 at the point   am a  ,  2 1  m2  1 m     a Here m = 1, a = a, so the required point is   2 , a   2 Circle Ex.17 Find the equation of tangents to the circle x2 + y2 – 6x + 4y – 12 = 0 which are parallel to the line 4x + 3y + 5 = 0 Sol. Given circle is x2 + y2 – 6x + 4y – 12 = 0 .......(i) and given line is 4x + 3y + 5 = 0 .......(ii) Centre of circle (i) is (3, –2) and its radius is 5. Equation of any line 4x + 3y + k = 0 parallel to the line (ii) ......(iii) If line (iii) is tangent to circle, (i) then | 4.3  3(2)  k | Point form of tangent : = 5 or |6 + k| = 25 2 2 4  3 2 2 2 (i) The equation of the tangent to the circle x + y = a or 6 + k = ± 25  k = 19, – 31 at its point (x1, y1) is, Hence equation of required tangents are x x1 + y y1 = a². 4x + 3y + 19 = 0 and 4x + 3y – 31 = 0 (ii) The equation of the tangent to the circle x2 + y2 + Equation of the Normal 2gx + 2fy + c = 0 at its point The equation of the normal at the point (x 1, y1) to (x1, y1) is: xx1 + yy1 + g (x+x1) + f (y+y1) + c = 0. the circle Note : In general the equation of tangent to any second x  x 1 y  y1 degree curve at point (x1, y1) on it can be obtained x2 + y2 + 2gx + 2fy + c = 0 is  by replacing x2 by x x1, y2 by yy1, x by x  x1 ,y 2 y  y1 x y  xy1 by , xy by 1 and c remains as c. 2 2 Parametric form of tangent : The equation of a tangent to circle x 2 + y2 = a2 at (a cos , asin ) is x cos  + y sin  = a. NOTE : The point of intersection of the tangents at the points P() & Q( ) is  a cos    a sin      2 2  ,      cos cos 2  2  Ex.16 Find the equation of the tangent to the circle x2 + y2 – 30x + 6y + 109 = 0 at (4, –1). Sol. Equation of tangent is x4   2  4x + (–y) – 30   y  ( 1)   2   +6  + 109 = 0 or 4x – y – 15x – 60 + 3y – 3 + 109 = 0 or –11x + 2y + 46 = 0 or 11x – 2y – 46 = 0 Hence, the required equation of the tangent is 11x – 2y – 46 = 0 x1  g y1  f x y Note : For the circle x2 + y2 = a2 it becomes x  y 1 1 Ex.18 Find the equation of the normal to the circle x2 + y2 – 5x + 2y – 48 = 0 at the point (5, 6). 5 2   Sol. Since normal is line joining centre  ,  1 and (5, 6) Slope =  14 5 Hence, the equation of the normal at (5, 6) is y – 6 = (14/5) (x – 5)  14x – 5y – 40 = 0 Length of the tangent The length of the tangent drawn from a point P(x1, y1) to the circle S  x 2 + y2 + 2gx + 2fy + c = 0 is given by PQ = PR = S1 where S1 = x12 + y12 + 2gx1 + 2fy1 + c Also area of the quadrilateral PQCR = r S1 and angle between tangents PQ and PR i.e.  QPR = 2tan–1 r S1 53 Circle Ex.22 A pair of tangents are drawn from the origin to the Pair of tangents from a point : circle x2 + y2 + 20 (x + y) + 20 = 0. Then find The equation of a pair of tangents drawn from the equation of the pair of tangent. point A (x1, y1) to the circle Sol. Equation of pair of tangents is given by SS1 = T2. x2 + y2 + 2gx + 2fy + c = 0 is : SS1 = T². or S = x2 + y2 + 20(x + y) + 20, S1 = 20, Where S  x2 + y2 + 2gx + 2fy + c ; T =10(x + y) + 20 = 0 S1  x1² + y1² + 2gx1 + 2fy1 + c  SS1 = T2 T  xx1 + yy1 + g(x + x1) + f(y + y1) + c.  20 {x2 + y2 + 20(x + y) + 20} = 102 (x + y + 2)2  4x2 + 4y2 + 10xy = 0  2x2 + 2y2 + 5xy = 0 Ex.19 Write the equation of the tangent to the circle (x – 1)2 + (y + 2)2 = 10 at the point (2, 1) Sol. The eqaution of the given circle can be written as x2 + y2 – 2x + 4y – 5 = 0. So the equation of the tangent at (2, 1) will be x(2) + y(1) – (x + 2) + 2(y + 1) – 5 = 0  x + 3y – 5 = 0 Ex.20 Find the equation of the normal to the circle x 2 + y2 + 6x + 8y + 1 = 0 passign through (0, 0) Sol. Centre of the circle = (–3, –4). So the normal is a line passing through (0, 0) and (–3, –4). Consequently its equation is y–0= 4 ( x  0) 3  4x – 3y = 0 Ex.21 Two tangents PQ and PR drawn to the circle x2 + y2 – 2x – 4y – 20 = 0 from point P(16, 7). If the centre of the cirlce is C then find the area of quadrilateral PQCR. 1 L×r 2 Ex.23 Find the equation of the pair of tangents drawn to the circle x2 + y2 – 2x + 4y = 0 from the point (0, 1) Sol. Given circle is S = x2 + y2 – 2x + 4y = 0 .......(i) Let P  (0, 1) For point P, S1 = 02 + 12 – 2.0 + 4.1 = 5 Clearly P lies outside the circle and T  x . 0 + y . 1 – (x + 0) + 2 (y + 1) i.e. T  –x +3y + 2. Now equation of pair of tangents from P(0, 1) to circle (1) is SS1 = T2 or 5 (x2 + y2 – 2x + 4y) = (– x + 3y + 2)2 or 5x2 + 5y2 – 10x + 20y = x2 + 9y2 + 4 – 6xy – 4x + 12y or 4x2 – 4y2 – 6x + 8y + 6xy – 4 = 0 or 2x2 – 2y2 + 3xy – 3x + 4y – 2 = 0 .......(ii) Separate equation of pair of tangents : From (ii), 2x2 + 3(y – 1) x – 2(2y2 – 4y + 2) = 0  x=  3( y  1)  9( y  1)2  8( 2y 2  4 y  2) 4 or 4x + 3y – 3 = ± 25 y 2  50 y  25 = ± 5(y – 1)  Separate equations of tangents are 2x – y + 1 = 0 and x + 2y – 2 = 0 DIRECTOR CIRCLE The locus of the point of intersection of two Where L = length of tangent and r = radius of circle. perpendicular tangents of a circle is called the director circle of that circle. L = S1 and r = 1  4  20  5 The equation of the director circle of x2 + y2 = a2 Hence the required area = 75 sq. units. is It may be easily seen that  Centre of the director circle = centre of the given circle.  Radius of the director circle = 2 (radius of the given circle) 54 Sol. Area PQCR = 2PQC = 2 × Circle Ex. 25 Find the distance between the chord of contact CHORD OF CONTACT : with respect to point (0, 0) and (g, f) of circle If two tangents PT1 & PT2 are drawn from the point 2 2 P(x 1, y1) to the circle S  x + y + 2gx + 2fy + c = 0, x2 + y2 + 2gx + 2fy + c = 0. then the equation of the chord of contact T1T2 is: Sol. Chord of contact with respect to (0, 0) xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0. gx + fy + c = 0 ....(i) Note : Here R = radius; L = length of tangent. Chord of contact with respect to (g, f) (a) Chord of contact exists only if the point ‘P’ is not gx + fy + g(x + g) + f(y + f) + c = 0 inside.  2gx + 2fy + g2 + f2 + c = 0 2 LR (b) Length of chord of contact T1 T2 = R 2  L2 .  gx + fy + 1 (g2 + f2 + c) = 0 ....(ii) 2 Distance betwen (i) and (ii) is 1 2 (g  f 2  c )  c g2  f 2  c 2  = g2  f 2 2 g2  f 2 Ex.26 Tangents are drawn to the circle x2 + y2 = 12 at (c) Area of the triangle formed by the pair of the tangents the points where it is met by the circle R L3 x2 + y2 – 5x + 3y – 2 = 0; find the point of & its chord of contact = R 2  L2 intersection of these tangents. (d) Tangent of the angle between the pair of tangents Sol. Given circles are S1  x2 + y2 – 12 = 0 ..... (i)  2RL  from (x1, y1) =  L2  R 2    (e) Equation of the circle circumscribing the triangle PT1 T2 is: (x  x1) (x + g) + (y  y1) (y + f) = 0. Ex.24 Find the equation of the chord of contact of the tangents drawn from (1, 2) to the circle x2 + y2 – 2x + 4y + 7 = 0 Sol. Given circle is x2 + y2 – 2x + 4y + 7 = 0 .......(i) Let P = (1, 2) For point P (1, 2), x 2 + y2 – 2x + 4y + 7 = 1 + 4 – 2 + 8 + 7 = 18 > 0 Hence point P lies outside the circle For point P (1, 2), T = x . 1 + y . 2 – (x + 1) + 2 (y + 2) + 7 i.e. T = 4y + 10 Now equation of the chord of contact of point P(1, 2) w.r.t. circle (i) will be 4y + 10 = 0 or 2y + 5 = 0 and S2 = x2 + y2 – 5x + 3y – 2 = 0 ...... (ii) Now equation of common chord of circle (i) and (ii) is S1 – S2 = 0 i.e. 5x – 3y – 10 = 0 ....... (iii) Let this line meet circle (i) [or (ii)] at A and B Let the tangents to circle (i) at A and B meet at P(), then AB will be the chord of contact of the tangents to the circle (i) from P, therefore equation of AB will be x + y – 12 = 0 ....... (iv) Now lines (iii) and (iv) are same, therefore, equations (iii) and (iv) are identical   5 =  3  = 12  10   = 6,  =– 18 5 18  Hence P =  6,  5    55 Circle POLE & POLAR Let P(x1, y1) be any point inside or outside the circle. Ex.27 Find the equation of polar of point ( 4, 4) with Draw chords AB and A' B' pasing through P. respect to circle ( x – 1)2 + ( y – 2)2 = 1. If tangent to the circle at A and B meet at Q (h, k), 2 2 then locus of Q is called polar of P.w.r.t. circle and P Sol. (x – 1) + ( y – 2) = 1 is called the pole and if tangent to the circle at A' and x2 + y2 – 2x – 4y + 4 = 0 B' meet at Q', then the straigt line QQ' is polar with equation of polar of point ( 4 , 4) is P' as its pole. 4x + 4y – (x + 4) – 2( y + 4) + 4 = 0  3x – 2y – 8 = 0 Ex.28 Find the pole of the line with respect to circle x2 + y2 = c2 . Sol. Let the pole is ( h , k) Hence polar of this pole is xh + yk – c2 = 0 .....(1) 1. Equation of polar  Equation of polar of the pole P(x1, y1) w.r.t. x2 + y2 = a2 is but polar is .....(2) comparing the coefficient of x and y xx1 + yy1 = a2  Equation of polar of the pole (x 1, y1) w.r.t. circle x2 + y2 + 2gx + 2fy + c = 0 is Equation of the chord with a given middle point: xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 2. Coordinates of pole  Pole of polar Ax + By + C = 0 w.r.t. circle x2 + y2 =  Aa 2 Ba 2 a is   C , C  2  The equation of the chord of the circle S  x2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M(x1, y1) is xx1 + yy1 + g (x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c which is designated by T = S1.     Pole of polar Ax + By + C = 0 with respect to circle x2 + y2 + 2gx + 2fy + c = 0 is given by the equation x 1  g y1  f gx 1  fy1  c   A B C 3. Conjugate points and Conjugate lines Notes : (i) (i) Conjugate points :- Two points are called conjugate points with respect to a circle if each point lies on the polar of the other point with respect to the same (ii) circle. (ii) Conjugate lines :- Two lines are called conjugate lines with respect to a circle if the pole of each line lies on the other line. The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose middle point is M. The chord passing through a point ' M ' inside the circle and which is at a maximum distance from the centre is a chord with middle point M. 56 Circle Ex.29 Find the equation of chord of the circle x2 + y2 = 8x bisected at the point ( 4 , 3). Sol. T = S1 x (4) + y(3) – 4 (x + 4) = 16 + 9 – 32 3y – 9 = 0 y=3 Ex.30 Find the equation of the chord of the circle x2 + y2 + 6x + 8y – 11 = 0, whose middle point is (1, –1) Sol. Equation of given circle is S  x2 + y2 + 6x + 8y – 11 = 0 Note :  Every Diameter passes through the centre of the circle.  A diameter is perpendicular to the system of parallel chords. Let L  (1, –1) For point L(1, –1), S1 = 12 + (–1)2 + 6.1 + 8(–1) – 11 = –11 and Ex.31 Find the diameter of the circle x2 + y2 – 4x + 2y – 11 = 0 corresponding to a system of chords parallel T  x.1 + y (–1) + 3(x + 1) + 4(y – 1) – 11 to the line x – 2y + 1 = 0. i.e. T  4x + 3y – 12 Sol. Centre of the circle = ( 2, –1). Now equation of the chord of circle (i) whose middle The equation of the line perpendicular to chord point is L(1, –1) is x – 2y + 1 = 0 is 2x + y + k = 0 T = S1 or 4x + 3y – 12 = –11 or 4x + 3y – 1 = 0 Since the line passes through the point ( 2, 1). So k = –3 Second Method : Let C be the centre of the given The equation of diameter is 2x + y – 3 = 0 circle, then C  (–3, –4). L  (1, –1) slope of CL 4  1 3 =  3 1 = 4  Equation of chord of circle whose middle point 4 is L, is y + 1 = – 3 (x – 1) ( chord is perpendicular to CL) or 4x + 3y – 1 = 0 FAMILY OF CIRCLES 1. If S = 0 and S' = 0 are two intersecting circles, the S +  S' = 0 (  –1) represents family of circles passing throguh intersection points of S = 0 and S' = 0 (  being parameter) Equation of the chord joining two points of circle : The equation of chord PQ to the circle x2 + y2 = a2 joining two points P() and Q() on it is given by the equation of a straight line joining two point  &  on the circle x2 + y2 = a2 is    x cos + y sin = a cos . 2 2 2 DIAMETER OF A CIRCLE 2. If S = 0 and L = 0 represent a circle and a line, then S + L = 0 represent family of circles passing through intersection points of circle S= 0 and line L = 0 (  being parameter) The locus of middle points of a system of parallel chords of a circle is called the diameter of that circle. The diameter of the circle x2 + y2 = r2 corresponding to the system of parallel chords y = mx + c is x + my = 0 57 Circle 3. Family of circles touching a line ax + by + c = 0 at Ex.34 Find the equations of the circles passing through (x1 , y1 ) on it, is the points of intersection of the circles 2 2 x2 + y2 –2x – 4y – 4 = 0 and x2 + y2 – 10x – (x – x1) + (y –y1) +  (ax + by + c) = 0 12y + 40 = 0 and whose radius is 4. (a) The equation of a family of circles passing through two given points (x1, y1) & (x2, y2) can be written in Sol. Any circle through the intersection of given circles is S 1 +  S2 = 0 or (x2 + y2 – 2x – 4y – 4) the form: + (x2 + y2 – 10x – 12y + 40 ) = 0 x y 1 (x  x1) (x  x2) + (y  y1) (y  y2) + K x1 y1 1 x2 y2 1 = 0, where K is a parameter. (b) Family of circles circumscribing a triangle whose sides are given by L1 = 0, L2 = 0 and L3 = 0 is given by; L 1L2 +  L2L3 +  L3 L1 = 0 provided co  efficient of xy = 0 and co  efficient of x2 = coefficient of y2. (c) Equation of circle circumscribing a quadrilateral whose side in order are represented by the lines L1 = 0, L2 = 0, L3 = 0 & L4 = 0 are u L1L3 +  L2L4 = 0 where values of u &  can be found out by using condition that coefficient of x2 = coefficient of y2 and coefficient of xy = 0. or (x 2 + y2) – 2 40  4 1  (1  5 ) 1  x – 2 =0 ( 2  6 ) 1  y + ...........(i) r = g2  f 2  c = 4, given  16 = (1  5 )2 (1   ) 2 + ( 2  6 )2 (1   ) 2 – 40   4 1  16(1 + 2 + 2) = 1 + 10 + 252 + 4 + 24 + 362 – 402 – 40 + 4 + 4 or 16 + 32 + 162 = 212 – 2 + 9 or 52 – 34 – 7 = 0  ( – 7) (5 + 1) = 0   = 7, – 1/5 Putting the values of  in (i) the required circles are 2x2 + 2y2 – 18x – 22y + 69 = 0 and x2 + y2 – 2y – 15 = 0 Ex.32 Find the equation of the circle pasing through the Ex.35 Find the equations of circles which touches origin and through the points of intersection of two 2x – y + 3 = 0 and pass through the points of circles x2 + y2 – 10x + 9 = 0 and x2 + y2 = 4 intersection of the line x + 2y – 1 = 0 and the circle Sol. Let the circle be (x 2 + y 2 – 10x + 9) + x2 + y2 – 2x + 1 = 0.  (x2 + y2 – 4) = 0 Sol. The required circle by S + P = 0 is Since it passes through ( 0,0), so we have x2 + y2 – 2x + 1 +  (x + 2y – 1) = 0 9 – 4 = 0  = 9/4 or x2 + y2 – x (2 – ) + 2y + (1 – ) = 0 So the required equation is 2  4(x2 + y2 – 10x + 9) + 9 ( x2 + y2 – 4) = 0 ,   centre (– g, – f) is  2   13(x2 + y2) – 40x = 0 1 (2   ) 2 Ex.33 Find the equation of the circle passing through the  2  (1   ) = r = g2  f 2  c = 2 4 origin and through the points of intersection of the 5 ||. circle x2 + y2 – 2x + 4y – 20 = 0 and the line 52 = 2 x + y –1 = 0 Since the circle touches the line 2x – y + 3 = 0 Sol. Let the required equation be therefore perpendicular from centre is equal to radius (x2 + y2 – 2x + 4y – 20) +  ( x + y – 1) = 0 2.((2   ) / 2)  ()  3 || = 5. Since it passes through ( 0,0), so we have –20 –  = 0 5 2   = –20  =±2 Hence the required equation is Putting the values of  in (i) the required circles are (x2 + y2 – 2x + 4y –20) –20 (x+y –1) = 0 x2 + y2 + 4y – 1 = 0 2 2 x + y –22x –16y = 0 x2 + y2 – 4x – 4y + 3 = 0. 58 Circle Ex.36 Find the equation of circle passing through the points A(1, 1) & B(2, 2) and whose radius is 1. Sol. Equation of AB is x – y = 0  equation of circle is (x – 1) (x – 2) + (y – 1) (y – 2) + (x – y) = 0 or x2 + y2 + ( – 3)x – ( + 3)y + 4 = 0 radius = 2 or coefficient of x2 = coefficient of y2  (   3)2 (  3) 2  4 4 4 5 + 3µ + 15 = 3 – 3 – µ 2 + µ + 3 = 0 ...........(ii) coefficient of xy = 0 But radius = 1 (given)  (5 + 3µ + 15)x2 + (3 – 3 – µ)y2 + xy (2 + 2µ – 14) – x (114 + 38 + 18µ) + y(50 – 14 – 6µ) + (207 + 69 + 27µ) = 0 ...........(i)  2 (  3 ) (  3 )  4 = 1 4 4 ( – 3)2 + ( + 3)2 – 16 = 4 + µ – 7 = 0 ..........(iii) Solving (ii) and (iii), we have =±1 2 2  equation of circles are x + y – 2x – 4y + 4 = 0 & x2 + y2 – 4x – 2y + 4 = 0 Ex.37 Find the equation of the circle passing through the point (2, 1) and touching the line x + 2y – 1 = 0 at the point (3, – 1).   = – 10, µ = 17  Puting these values of  & µ in equation (i), we get 2x2 + 2y2 – 5x + 11y – 3 = 0 COMMON CHORD OF TWO CIRCLES (x – 3)2 + (y + 1)2 + (x + 2y – 1) = 0 The line joining the points of intersection of two circles is called the common chord. If the equation of two circle. Since it passes through the point (2, 1), S Sol. Equation of circle is 1 + 4 +  (2 + 2 – 1) = 0  =– 5 3  circle is (x – 3)2 + (y + 1)2 –  3x2 + 3y2 – 23x – 4y + 35 = 0 5 3 (x + 2y – 1) = 0 Ex.38 Find the equation of circle circumcscribing the triangle whose sides are 3x – y – 9 = 0, 5x – 3y – 23 = 0 & x + y – 3 = 0. Sol. x2 + y2 + 2g1x + 2f1y + c1 = 0 and S' x2 + y2 + 2g2x + 2f2y + c2 = 0 , then equation of common chord is S – S' = 0 2(g1 – g2)x + 2 (f1 – f2)y + c1 –c2 = 0 Also the length of the common chord AB is given by AB  2 a 2  p 2 where 'a' is the radius of one of the given circles and 'p' is the distance of its centre from their common chord. Ex.39 Find the length and equation of the common chord of circles x2 + y2 = 10x and x2 + y2 = 4 Sol. The equation of the common chord is L1L2 + L2L3 + µL1L3 = 0 (3x – y – 9) (5x – 3y – 23) + (5x – 3y – 23) (x + y – 3) + µ (3x – y – 9) (x + y – 3) = 0 (15x 2 + 3y2 – 14xy – 114x + 50y + 207) + (5x2 – 3y2 + 2xy – 38x – 14y + 69) + µ (3x2 – y2 + 2xy – 18x – 6y + 27) = 0 (x2 + y2 – 10x) – (x2 + y2 – 4) = 0 – 10x + 4 = 0 5x – 2 = 0 Also with respect to second circle a = 2 , p = 2/5 length of common chord = 59 Circle Sol. Let the two circles be x + y + 2g1 x + 2f1 y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 , 2 ANGLE OF INTERSECTION OF TWO CIRCLES The angle of intersection between two circles S = 0 and S' = 0 is defined as the angle between their tangent at their point of intersection. 2 where g1 = 5/2, f1 = 3/2 , c1 = 7, g2 = –4, f2 = 3 and c2 = m. If the two circles intersects orthogonally, then 2( g1 g2 + f1 f2 ) = c1 + c2 11 = 7 + m A =7+m m = –18 If S  x2 + y2 + 2g1x + 2f1y + c1 = 0 Ex.41 Obtain the equation of the circle orthogonal to both the circles x2 + y2 + 3x – 5y+ 6 = 0 and 4x2 + 4y2 – 28x +29 = 0 and whose centre lies on the line 3x + 4y + 1 = 0. S'  x2 + y2 + 2g2x + 2f2y + c2 = 0 Sol. Given circles are x2 + y2 + 3x – 5y + 6 = 0 .....(i) are two circles with radii r1, r2 and d be the distance between their centres then the angle of intersection  between them is given by cos = 2r1 r2 and 4x2 + 4y2 – 28x + 29 = 0 or x2 + y2 – 7x + ..........(ii) Let the required circle be x 2 + y2 + 2gx + 2fy + c or = 0 ..........(iii) Since circle (iii) cuts circles (i) and (ii) orthogonally cos = Condition of Orthogonality If the angle of intersecton of the two circle is a right angle then such circle are called Orthogonal circle. In PC1C2  d2 = r12 + r22 (g1–g2)2 + (f1–f2)2 = g12+f12–c1+g22 + f22 – c2  2g1 g2 + 2f1f2 = c1 + c2  (Condition of Orthogonality) 3  5 2g  2  + 2f   2  = c + 6 or     3g – 5f = c + 6 ...........(iv)  (C1C2)2 = (C1P)2 + (C2P)2 29 = 0. 4  7 and 2g   2  + 2f.0 = c +  – 7g = c +  29 4 29 4 or ...........(v) From (iv) & (v), we get 10g – 5f = – 5 4 or 40g – 20f = – 5. ..........(vi) Given line is 3x + 4y = – 1 ..........(vii) Since centre (– g, – f) of circle (iii) lies on line (vii),  – 3g – 4f = – 1 .........(viii) Solving (vi) & (viii), we get g = 0, f = Ex.40 For what value of m the circles x2 + y2 + 5x + 3y + 7 = 0 and x 2 + y2 – 8x + 6y + m = 0 cuts orthogonally 29 4  from (5), c = –  from (iii), required circle is x2 + y2 + 1 4 1 29 y– =0 2 4 or 4(x2 + y2) + 2y – 29 = 0 60 Circle POSITION OF TWO CIRCLES Let C1 (h1 , k1) and C2 ( h2, k2) be the centre of two circle and r1, r2 be their radius then Point of intersection of common tangents : The points T1 and T2 (points of intersection of indirect and direct common tangents) divide C1C2 internally and externally in the ratio r1 : r2. Equation of the common tangent at point of contact : S1 – S2 = 0 Point of contact : The point of contact C1C2 in the ratio r1 : r2 internally or externally as the case may be. (ii) Length of an external (or direct) common tangent & internal (or transverse) common tangent to the two circles are given by: Lext = d2  (r1  r2 )2 & Lint = d2  (r1  r2 )2 , where d = distance between the centres of the two circles and r1, r2 are the radii of the two circles. Note that length of internal common tangent is always less than the length of the external or direct common tangent. Notes : (i) The direct common tangents meet at a point which 2 2 divides the line joining centre of circles externally in Ex.42 Examine if the two circles x + y – 2x – 4y = 0 and x2 + y2 – 8y – 4 = 0 touch each other externally the ratio of their radii. or internally. Transverse common tangents meet at a point which 2 2 divides the line joining centre of circles internally in Sol. Given circles are x + y – 2x – 4y = 0 ...........(i) and x2 + y2 – 8y – 4 = 0 ...........(ii) the ratio of their radii. 61 Circle Let A and B be the centres and r1 and r2 the radii of circles (i) and (ii) respectively, then A  (1, 2), B  (0, 4), r1 = 5, r2 = 25 Now AB = (1  0) 2  ( 2  4)2 = 5 and r1 + r2 = 3 5 , |r1 – r2| = 5 Thus AB = |r1 – r2|, hence the two circles touch each other internally. Ex.43 A circle with radius 5 touches another circle x2 + y2 – 2x – 4y – 20 = 0 at point (5, 5). Find its eqaution. Sol. The centre of the given circle C1  (1, 2) and radius = 1  4  20  5 . Since the radii of two circles are Notes : (a) If two circles intersect, then the radical axis is the common chord of the two circles. (b) If two circles touch each other, then the radical axis is the common tangent of the two circles at the common point of contact. (c) Radical axis is always perpendicular to the line joining the centres of the two circles. (d) Radical axis will pass through the mid point of the line joining the centres of the two circles only if the two circles have equal radii. (e) Radical axis bisects a common tangent between the two circles. (f) A system of circles, every two which have the same radical axis, is called a coaxial system. (g) Pairs of circles which do not have radical axis are concentric. equal so they touch externally. If C2(h, k) be the centre of the required circle then the point of contact (5, 5) is the mid point of C1C2. Hence C2  (9, 8) Ex.44 Find the co-ordinates of the point from which the lengths of the tangents to the following three circles and the reqd. eqn will be (x – 9)2 + (y – 8)2 = 25 be equal.  x2 + y2 – 18x – 16y + 120 = 0 a 3x2 + 3y2 + 4x – 6y – 1 = 0 2x2 + 2y2 – 3x – 2y – 4 = 0 Radical axis and radical centre: 2x2 + 2y2 – x + y – 1 = 0 The radical axis of two circles is the locus of points Sol. Here we have to find the radical centre of the three circles. First reduce them to standard form in which whose powers w.r.t. the two circles are equal. coefficients of x2 and y2 be each unity. Subtracting in The equation of radical axis of the two circles S1 = 0 pairs the three radical axes are & S2 = 0 is given by 17 5 3 3 x–y+ =0 ; –x– y– =0 6 3 2 2 S1  S2 = 0 i.e. 2(g1  g2) x + 2(f1  f2) y + (c1  c2) = 0. 11 5 1 – 6 x + y – 6 = 0. 2  16 31  solving any two, we get the point   21  63  which   satisfies the third also. This point is Ex.45 Find the equation opf the radical axis of the circles 2x2 + 2y2 – 7x = 0 and x2 + y2 – 4y – 7 = 0. Sol. The equations of the given circles may be written as 2x2 + 2y2 – 7x = 0 and 2x2 + 2y2 – 8y – 4 = 0 The common point of intersection of the radical axes The equation of their radical axis is given by of three circles taken two at a time is called the radical S – S' = 0 centre of three circles. Note that the length of tangents  (2x2 + 2y2 – 7x) – (2x2 + 2y2 – 8y – 14) = 0 from radical centre to the three circles are equal.  7x – 8y – 14 = 0 62  SOME IMPORTANT RESULTS      x If the circle x2 + y2 + 2gx + 2fy + c = 0 is a point circle then g2 + f2 = c. 2  The length of the tangent drawn from any point on the circle x2 + y2 + 2gx + 2fy + c1 = 0 to the circle  x2 + y2 + 2gx + 2fy + c = 0 is c  c 1 . 1 1  If two tangents drawn from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0 are perpendicular to each other, then g2 + f2 = 2c.  If the line y = mx + c is a normal to the circle with radius r and centre at (a, b), then b = ma + c.  If the tangent to the circle x2 + y2 = r2 at the point (a, b) meets the coordinates axes at the points A and B and O is the origin, then the area of the triangle OAB is 2 2 –1  a   2 2 2    a    .  If lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 cut the axes at concyclic points, then l1l2 = m1m2. The area of the triangle formed by the tangents from the points (h, k) to the circle x2 + y2 = a2 and their a 2 h  k2 (h2  k 2  a 2 ) 3 / 2 . If O is the origin and OP, OQ are tangents to the circle x2 + y2 + 2gx + 2fy + c = 0, then the g f  2ab . The angle between the tangents from (, ) to the chord of contact is The length of the common chord of the circles   circumcentre of the triangle OPQ is  2 , 2  .   If chord of a circle AB and CD meet at some point P, then PA . PB = PC . PD and if AB chord and tangent at T meet at P, then PA . PB = PT2. r4 . 2ab If  is the angle subtended at P(x1, y1) by the circle S = x2 + y2 + 2gx + 2fy + c = 0, then  S 1 cot 2  2 2 g f c    a2  b 2 1 g2  f 2  c 2 g2  f 2 . circle x + y = a is 2 tan The length of the common chord of the circles x2 + y2 + ax + by + c = 0 and x2 + y2 + bx + ay + c = 0 1 (a  b)2  4c 2     The distance between the chord of contact of the tangents to the circle x2 + y2 + 2gx + 2fy + c = 0 from the origin and the point (g, f) is 2 If the circles x2 + y2 + 2gx + c2 = 0 and x2 + y2 + 2fy (x – a)2 + y2 = a2 and x2 + (y – b)2 = b2 is   If the radius of the given circle x + y + 2gx + 2fy + c= 0 be r and it touches both the axes then g = f = c = r.. is   r 2 (a 2  b 2 )  a 2b 2 2  a2  b2  2 + c2 = 0 touch each other, then g2  f 2  c 2 . If the circle x2 + y2 + 2gx + 2fy + c = 0 touches x-axis and y-axis, then g2 = c and f2 = c respectively. y x2 + y2 = r2 on the line a  b  1 is If the line lx + my + n = 0 is a tangent to the circle x2 + y2 = a2, then a2 (l2 + m2) = n2. 1  Circle The length of the chord intercepted by the circle If the line lx + my + n = 0 is a tangent to the circle (x – h)2 + (y – k)2 = a2, then (hl + km + n)2 = a2 (l2 + m2).  If OA and OB are the tangents from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0 and C is the centre of the circle then the area of the quadrilateral OABC is c(g 2  f 2  c ) . 63