Circle Exercise # 3

EXERCISE # 3 Part-A Subjective Type Questions Q.1 Q.3 Find the equations of straight lines which pass through the intersection of the lines x – 2y – 5 = 0, 7x + y = 50 & divide the circumference of the circle x2 + y2 = 100 into two arcs whose lengths are in the ratio 2 : 1. Sol. The equation of line passing through the point of intersection of the lines x – 2y – 5 = 0 and 7x + y – 50 = 0 is given by Find the locus of the middle points of chords of circle x2 + y2 = a2 which subtend a right angle at the point (c, 0). y A Sol. N (h,k) x' P O (c,0) (x – 2y – 5) + (7x + y – 50) = 0 x  x(1 + 7) + y ( – 2) – 50 – 5 = 0 B A y' Let N (h, k) be the middle point of any chord AB, which subtends a right angle at P(c,0) since APB = 90º  NA = NB = NP (since distance of the vertices from middle point of the hypotenuse are equal) or (NA)2 = (NB)2 = (h – c)2 + (k – 0)2 .......(i) But also BNO = 90º  (OB)2 = (ON)2 + (NB)2 – (NB)2 = (ON)2 – (OB)2  – [(h – c)2 + (k – 0)2] = (h2 + k2) – a2 or 2(h2 + k2) – 2ch + c2 – a2 = 0  Locus of N(h, k) is 2(x2 + y2) – 2cx + c2 – a2 = 0 Q.2 Sol. .........(i) O(0,0) 60º C 60º B Suppose this line cuts the given circle at A and B such that arc APB : arc AQB = 2 : 1  AOB = 120º  AOC = BOC = 60º – 50 – 5 Now,OC= 50  5 = 502  10  5 (1  7 ) 2  ( – 2) 2 and, OA = (radius of the circle x2 + y2 = 100)  OA = 10 In AOC, we have cos60º = A variable circle passes through the point A (a, b) & touches the x-axis ; show that the locus of the other end of the diameter through A is (x – a)2 = 4by. OC OA  OC = OA cos 60º  OC = 10. Let the other end of the diameter through pa qb , A(a,b) be B(p,q) so that centre is   2   2 and (diameter)2 = 4r2 = (p–a)2 + (q – b)2 ....(i) Since the circle touches x-axis, its radius is qb ordinate of centre i.e. =r 2 or (q + b)2 = 4r2 ..........(ii) 2 2  (q + b) –(q – b) = (p – )2 [from (i) & (ii)] or (p – q)2 = 4bq Hence locus of (p, q) is (x – a)2 = 4by  1 =5 2 50  5 502  10  5 =5  (10 + 1)2 = 502 + 10 + 5  25 2 + 5 – 2 = 0  (5 + 2) (5 – 1) = 0   = 1 –2 , 5 5 Putting the values of  in (i), we get the reqd. lines 4x – 3y = 25 and 3x + 4y = 25 Q.4 Find an equation of the circle which touches the straight - lines x + y = 2, x – y = 2 and also touches the circle x2 + y2 = 1 externally. 1= 0 Sol. C2 (0,0) C1 (h,0) Q.6 Find the equation of the smallest circle passing through the intersection of x2 + y2 – 2x – 4y – 4 = 0 and the line x + y – 4 = 0. Sol. Any circle passing through the points of intersection of the given circle and the line has the equation as A(2,0) 2= 0 1 = 0 & 2 = 0 are tangents to required circle which intersect at A(2, 0). The centre of the required circle will be lie on bisector of acute angle between these tangnets xy–2 x–y–2 i.e. on = i.e. y = 0 2 2 let it be (h, 0) where h is +ve and if r be the radius then it toueches x2 + y2 = 1 externally. C1C2 = r1 + r2 or h = 1 + r form (i) h–2 P = r with any tangent gives =r 2 (x2 + y2 – 2x – 4y – 4) +  (x + y – 4) = 0  –2 –4 ,– its centre is  –  2 2   The circle is the smallest if its certre lie on the given line (chord) –  =4– 2,  2 = – 2   = –1 Putting  = –1 in equation (i), weget x2 + y2 – 3x – 5y = 0 Hence x2 + y2 – 3x – 5y = 0 2 and radius ( 2 – 1) Q.7  (x – 4 + 2 )2 + y2 = ( 2 – 1)2 and (x – Q.5 Sol. 2 )2 + y2 = ( 2 – 1)2 A tangent is drawn to each of the circle x2 + y2 = a2 and x2 + y2 = b2. Show that if the two tangents are mutually perpendicular, the locus of their points of intersection is a circle concentric with the given circles. Any tangent to first circle (x2 + y2 = a2) is given by x cos + y sin  = a As its distance from centre (0, 0) is equal to radius a any tangent to x2 + y2 = b2 but perpendicular to above is obtained by replacing  by  – 90º and its equation is xcos( – 90º) + ysin ( – 90º) = b or x cos(90º – ) – y sin (90º – ) = b or x sin – y cos = b locus of the point of intersection of these tangents is x2 + y2 = a2 + b2 which is a circle concentric with the given circles. 2– 4– + –4=0 2 2  – 2 – 2 = 0 2 ( 2 – 1)  centres of circles are (4 – 2 , 0),( 2 , 0) –2 –4 – –4=0 2 2 2–+4––8=0  h = 2 ± 2 r = 1 + r form (i) 1 1 = (1 + 2 )r  r = = 2 –1 1 2 Only as r = + ve h=2± .........(i) A point A (2, 1) is outside the circle x2 + y2 + 2gx + 2fy + c = 0 & AP, AQ are tangents to the circle. Find the equation of the circumcircle of the triangle APQ. P Sol. A (2,1) C (–g,–f) Q centre of the required circle is the mid-point of A(2, 1) and centre C of the given circle  2 – g 1– f  ,    is the centre of required circle 2   2 2 and radius = 2–g  1– f  – 2   – 1   2   2  = 1 2 (g  2)2  (f  1)2 = 1 2 g 2  4  4g  f 2  2f  1  Reqd. circle is 2 2 2–g  1– f   x –  +y –  2 2     =  2  2m m( 2  2m)  ,  P is   1  m2   1  m2 2 P (g 2  f 2  2f  4g  5) 4 [2x–(2–g)]2 + [2y–(1–f)]2 = g2 + f2 + 2f + 4g + 5 4x2+(2 – g)2– 4x (2 – g)+4y2 – 4y (1 – f)+(1– f)2 = g2 + f2 + 2f + 4g + 5 4x2 + 4y2 – 8x – 4y + 4xg + 4fy–2f–4g=0 O   C (1,1) Q  x2 + y2 – 2x – y – xg – fy = 0  x2 + y2 + x(g – 2) + y (f – 1) – 2g – f = 0 Q.8 Sol. Since QQ is equally inclined to x-axis  tan (– ) = – tan = – m If S1  x2 + y2 + 2g1x + 2f1y + c1 = 0 and S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 are two circles with radii r1 and r2 respectively, show that the points at which the circles subtend S S equal angles lie on the circle 21 = 22 r2 r1 tan =  2 – 2m – m( 2 – 2m)  ,  Point Q is   1  m2   1  m2 If (h, k) be the mid point of PQ then h= 2m 2 1  m2 Q.10 The centre of the circle S = 0 lie on the line 2x – 2y + 9 = 0 & S = 0 cuts orthogonally the circle x2 + y2 = 4. Show that circle S = 0 passes through two fixed points & find their coordinates. Sol. Let circle be S  x2 + y2 + 2gx + 2fy + c = 0 ..(i) since centre of this circle (–g,–f) lie on 2x – 2y + 9 = 0 S S  21 = 22 r1 r2  – 2g + 2f + 9 = 0 and the circle S = 0 and x2 + y2 – 4 = 0 cuts orthogonally Find the locus of the mid point of all chords of the circle x2 + y2 – 2x – 2y = 0 such that the pair of lines joining (0, 0) & the point of intersection of the chords with the circles make equal angle with axis of x.  2g × 0 + 2f(0) = c – 4 c=4 ........(iii) Substituting the values of g and c from (ii) and (iii) in (i) then The given circle is x2 + y2 – 2x – 2y = 0  x2 + y2 + (2f + 9) x + 2fy + 4 = 0 or (x2 + y2 + 9x + 4) + 2f (x + y) = 0 Hence the circle S = 0 passes through fixed point  x2 + y2 +9x + 4 = 0 and x + y = 0 After solving, we get  (x – 1)2 + (y – 1)2 = 2 Whose centre is (1, 1) and radius equal to 2 it passes through origin (0, 0) i.e. O(0, 0) Let OP be any chord through origin whose equation is y = mx where m = tan solving with circle x2 + y2 – 2x – 2y = 0 we get x2 + m2x2 – 2x –2y = 0  x(x +m2x – 2 – 2m) = 0  x = 0 and x = 1  m2 and k =  locus is x + y = 2  r22 S1 – r12 S2 = 0 Sol. 2 h+k=2 r1 r = 2 t1 t 2  t1r1 = t2r1 on squaring t12 r22 = t22 r12 or S'2r12 = S'1 r22 = 0 on generalisation Q.9 A 2  2m 1  m2  1 1 (–4, 4) or  – ,   2 2 Q.11 Show that the common tangents to the circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0 form an equilateral triangle. 25 Sol. Here the circle x2 + y2 – 6x = 0 has the centre = (3, 0) and radius=3 and the circle x2+y2 + 2x = 0 has the centre = (–1, 0) & Now we have to prove that lines x=0,y= (3  1) 2  02 = 4 radius = –x 3 – 3 and y = x + 3 3 Forms an equilatarl triangle y x=0 y= x  3 and sum of radii = 3 + 1 = 4  the circles touch externally 3 x (–1,0) 3 y = –x – 3 (3,0) 3 Clearly, the indination of y = obviously, the point of contect = (0,0) let y = mx + c be a common tangent to the circle except at the point of contact. Then the length of the perpendicular from (–1, 0) to y = mx + c is equal to the radius of second circle  –mc 1  m2 3m  c 1  m2 ......(ii) from (i)  c2– 2mc = 1 ......(iii) form (ii)  6mc + c2 = 9 .......(iv)  (iii) given c = 3 ,  c = ± 2 The centres of similitud from where the common tangents pass are easily found to be the points (15/2, 0) and (30, 0)  Required four common tangents are given by 5 y = 2x – 15, Q.13 circumference is xx1 + yy1 – a2 + 3 Sol. 3 from (iii), weget 1 3 d2 = 0. 2 Let R (x1,y1) be a point on the circle x2 + y2 = a2  x12 + y12 = a2 .........(i) R(x1,y1) 3 d a 1 P 3  two common tangents are x+ 5 y = – 2x + 15 35 y = x – 30, 35 y = – x + 30 Show that the equation of a straight line meeting the circle x2 + y2 = a2 in two points at equal distances 'd' from a point (x1, y1) on its when c = – 3 form (iii), we get 1  as 6 Sol. or 8 mc = 8  mc = 1 y=  3 is Find the equation to the four common tangents to the circles x2 + y2 = 25 and (x –12)2 + y2 = 9.  (6mc+ c2) – (c2 – 2mc) = 9 – 1 = 8 3 + 2m 3 = 1  m = – 3 Q.12 =3 3 – 2m 3 = 1  m = x – 3 with the  we get from the figure that the triangle is equilateral. or (3m + c)2 = 9 (1+m2) when c = 3 shown in Also their point of intersection is (–3, 0) =1 or (–m + c)2 = 1+ m2 ..............(i) and the lenght of the perpendicular from (3, 0) to y = mx + x is equal to the radius of the first circle  5 and that of y = 6 x-axis is –x 3 and y = – 1 3 x– 3 The tangent at the point of contanct (0, 0) is x.0 + y.0 + (x + 0) = (x + 0) = 0  x = 0 a d O Q Let P and Q be two points on the given circle such that PR = QR = d Now line joining R and O (the centre of the circle) is perpendicular to PQ and meets PQ at L (say).  let equation of PQ be – x1 x+c y1 or xx1 + yy1 = cy1 y= – cy1  OL = and RL = .............(ii) x12  y12 – cy1 a = x12  y12 – cyl x12 = a–  y12 cy1 a cy   PL2 = d2 –  a – 1  a   from RPL, 2 2  cy  from OPL, PL2 = a2 –  1   a  Equating the two values, we get 2 cy   d 2 –  a – 1  = a2 – a    cy1     a  ` 2 2 2  cy   cy  = d2 – a2 + 2cy1 –  1  = a2 –  1  a    a  2 2 2  d – a + 2cy1 = a  2cy1 = 2a2 – d2 d2 2  from (ii), we get the requird equation of the line as Q.15 From a point P tangent drawn to the circles x2 + y2 + x – 3 = 0, 3x2 + 3y2 – 5x + 3y = 0 & 4x2 + 4y2 + 8x + 7y + 9 = 0 are of equal lengths. Find the equation of the circle through P which touches the line x + y = 5 at the point (6, –1). Sol. Here point P is the radical centre of these circles. On solving like S1 – S2 = 0 & S2 – S3 = 0 we get point P(0, –3). Given reqd. circle passes through P and touches x + y = 5 at point (6, –1) (0,–3) (, k) r  cy1 = a2 – d2 d2 xx1 + yy1 = a –  xx1 + yy1 – a2 + =0 2 2 2 Q.14 Sol. Through the point of intersection P which has integral coordinates of the circles x2 + y2 = 1 and x2 + y2 + 2x + 4y + 1 = 0, a common chord APB is drawn terminating on the two circles such that the chords AP and BP of the given circles subtend equal angles at the centre. Find the equation of this chord. (6, –1) x + y = 5 Let centre of reqd. circle is (h, k) The given circle intersect at the point (–1, 0) and (3/5, –4/5).since P has integral co-ordinates, the co-ordinates of P are (–1, 0) A (–1,0)P    C1(0,0)  B C2(–1,–2) let an equation of the chord AB through P be y = m(x + 1) let AP and BP make the same angle 2a at the centres C1(0, 0) and C2(–1,–2) respectively of the gieven circles.  APC1 = BPC2 = – ..........(i) 2 Now slope of C1P = 0 and slope of C2P = 2/m' where m' = 0 from (i), we have 2 m– m–0   m tan  –   = = 2 1  m.0 2  1  m. m mm'– 2 1  |m| = = m '2 m m  m2 = 1  m = ± 1 so that the required equation of the chord is y = x + 1 corresponding to the value m = 1 (for m = –1, the equation will represent the chord PAB, A and B lying on the same side of P, Hence we donote consider it).  h 2  (k  3) 2  (h  6) 2  (k  1) 2  h2 + k2 + 6k + 9 = h2 – 12h + 36 + k2 + 2k + 1  3h + k = 7 ……..(1) Now equation of line perpendicular to x + y = 5 is given by x – y =  since it passes (6, –1) then 6 + 1 =  = 7 x–y=7 this also passes (h, k)  h – k = 7 ……..(2) from (1) & (2), we get 4h = 14 7 7 h = 7/2 k = – 7 = – 2 2 equation of C2 is (x – h)2 + (y – k)2 = 12 2 2 x + y – 2hx – 2ky + h2 + k2 – 1 = 0  PA1 is tangent to C2 | 4h  9 |  =1 5 95 7 h= = 1 or 4 2 7  h= (as (1, 1) is centre of another circle) 2 7 7 (h, k)   ,  2 2 2 radius r = 7 7     3   2 2    2 49 1 50 =   4 4 4 7 7 since centre (h, k) =  ,  2 2 = Q.16 50 4 Equation of required circle is given by radius r = 2 2  50  7  7    x     y      2  2 4    2 2 Sol.[C] Angle subtended by circle C1 at P is r 2 tan–1 S1 2 7  7 50   x   y    2  2 4   98 50   x2 + y2 – 7x + 7y +    = 0 4   4 2 2  x + y – 7x + 7y + 12 = 0  x2 + y2 – 7x + 7y + 12 = 0 = 2 tan–1 Q.17 Part-B Passage based objective questions Passage I (Question 16 to 18) Let C1, C2 are two circles each of radius 1 touching internally the sides of triangles POA1, PA1A2 respectively where P  (0, 4), O is origin, A1, A2 are the points on positive x-axis. On the basis of above passage, answer the following questions : Y Sol. (0,4) 1 (h,k) 1 1 1 C2 A1 1 2/3 3 = tan–1  tan 1   2 3 1  (1 / 3) 4 Centre of circle C2 is(A) (3, 1) (C) (3 (B) (3 3 , 1) 4 1 , 1) 2 (D) None of these  1  Sol.[B] Centre of circle C2 is  3 ,1  2  Q.18 Slope of line PA2 is–3 –8 –4 (A) (B) (C) 3 4 15 (D) –2 3 Sol.[C] Slope of PA2 = ? equation of tangent from (0, 4) to the circle C2 is 7 y – 1 = m (x – ) + 1 m 2 2  (0, 4) satisfies this equation P C1 1 Angle subtended by circle C1 at P is2 2 (A) tan–1 (B) 2 tan–1 3 3 3 3 (C) tan–1 (D) 2 tan–1 4 4 A2 C1  (x – 1)2 + (y – 1)2 – 1 = 0 x2 + y2 – 2x – 2y + 1 = 0 let equation of PA, is y – 4 = mx  PA1 is tangent to C1 | m3|  =1 1 m2  m2 + 6m + 9 = 1 + m2 4  m =  and – 3  equation of PA1 is 4x + 3y = 12 X  7 3 = m     1 m2  2 m=  4 8 or  3 15 Passage II (Question 19 to 21) A line 2x + y = 3 intersects circle x2 + y2 – 4x – 6y + 4 = 0 at A and B. Tangents at A and B are drawn which meets at P. M is the mid point of AB On the basis of above passage, answer the following questions : Q.19 (A) x2 + y2 – 3x – 3y – 5 = 0 (B) x2 + y2 – 3x – 3y – 4 = 0 (C) x2 + y2 + 3x + 3y – 4 = 0 (D) x2 + y2 + 3x + 3y – 5 = 0 Length of chord AB is equal to29 5 (A) 2 29 29 (C) (D) 5 5 (B) 2 29 5 S Sol.[B] Sol.[B] r=3 A (2,3) Op M B P Equation of circle passing through AB is given by S1 + L = 0 (or S2 + L = 0) R i.e., x2 + y2 – 2x – 2y – 7 + (x + y – 3) = 0 i.e., x2 + y2 + ( – 2)x + ( – 2)y – (7 + 3) = 0 5 Q.20 16 29 = 2 5 5 Co-ordinate of M is 2 11  (A)  ,  5 5  3 9 (C)  ,  5 5   (   2)  (   2 )  , C    2 2    AB is a diameter of this circle C lies on AB  – ( – 2) = 3  = – 1  equation of the circle is x2 + y2 – 3x – 3y – 4 = 0  1 13  (B)  ,  5 5  (D) None of these Sol.[A] Equation of AB is L = 2x + y = 3 equation of OP is x – 2y = – 4 intersection point of AB and OP is  2 11  M  ,  5 5  Q.21 S2 Q B 4   = 2 32  A P length of chord() = 2 r 2  p 2 p= S1 Equation of circum circle of triangle PAB is(A) 4x2 + 4y2 + 4x – 15y – 11 = 0 (B) 4x2 + 4y2 + 2x – 15y – 11 = 0 (C) 4x2 + 4y2 – 2x – 15y – 11 = 0 (D) 4x2 + 4y2 – 4x – 15y – 11 = 0 Sol.[B] Equation of circle passing through intersection point of the circle S  0 and L  0 is S + L = 0 7 4  equation is 4x2 + 4y2 + 2x – 15y – 11 = 0  P satisfies it  = Passage III (Question 22 to 24) Two circles S1 : x2 + y2 – 2x – 2y – 7 = 0 and S2 : x2 + y2 – 4x – 4y – 1 = 0 intersects each other at A and B. Tangents at A & B to circle S2 meet at Q. On the basis of above passage, answer the following questions : Sol Equation of common chord is x+y=3 Q.22 Equation of circle passing through A and B whose AB is diameter- Q.23 Co-ordinates of point Q are(A) (–6, –6) (B) (–5, –5) (C) (–7, –7) (D) None of these Sol.[C] Let Q  (h, k) then equation of AB is hx + ky – 2(x + h) – 2 (y + k) – 1 = 0 i.e., (h – 2) x + (k – 2) y – (2h + 2k + 1) = 0 comparing this equation with x + y – 3 = 0 gives Q  (h, k) = (–7, –7). Q.24 Mid point of AB is5 1 (A)  ,  2 2 (C) (2, 1) 3 3 (B)  ,  2 2 (D) (1, 2) Sol.[B] Let (a, b) be mid point of AB then equation of AB (w.r.t. S1 = 0) is T = S1 i.e., ax + by – (x + a) – (y + b) – 7 = a2 + b2 – 2a – 2b – 7 Comparing this equation with x + y = 3 gives 3 3 M  (a, b)   ,  2 2 EXERCISE # 3 1. 2(x2 + y2) – 2cx + c2 – a2 = 0 3. 4x – 3y – 25 = 0 ; 3x + 4y = 25 4. (x + 4 + 3 2 )2 + y2 = 9( 2 + 1)2, (x – 2 )2 + y2 = (–1 + 2 )2 5. x2 + y2 = a2 + b2 6. x2 + y2 – 3x – 5y = 0 7. x2 + y2 + (g – 2) x + (f – 1)y – 2g – f = 0 9. x + y = 2 10. (–4, 4), (– 12. 2x ± y 5 – 15 = 0 or x ± y 35 – 30 = 0 14. y = x + 1, y + x + 1 = 0 1 1 , ) 2 2 15. x2 + y2 – 7x + 7y + 12 = 0 Qus. 16 17 18 19 20 21 22 23 24 Ans. C B C B A B B C B