Circle Exercise # 2

EXERCISE # 2 Part-A Only single correct answer type questions Q.1 If a line is drawn through a fixed point Sol. [B] Let S = x2 + y2 – a2 = 0 Q.3 5 of the circle is – (A) (2, 29/10) (B) (29/10, 2) (C) (– 2, 29/10) (D) None of these Sol. [B] 4 have roots , 5 5 4 29  + 5 = – 5 10 Q.4 The equation of the circle of radius 22 whose centre lies on the line x – y = 0 and which touches the line x + y = 4, and whose centre's coordinates satisfy the inequality x + y > 4 is (A) x2 + y2 – 8x – 8y + 24 = 0 (B) x2 + y2 = 8 (C) x2 + y2 – 8x – 8y = 24 (D) None of these Sol. [A] Let centre  (t, t) T (,) P A B  PA.PB = PT2 where PT is length of tangent  PA.PB =  s1 2  PA.PB = s1  PA.PB = 2 + 2 – a2 Q.2 Sol. , and are the parametric angles of three points P, Q and R respectively on the circle x2 + y2 = 1 and A is the point (–1, 0). If the lengths of the chords AP, AQ and AR are in    G.P., then cos , cos and cos are in 2 2 2 (where , ,   (0, )) (A) A.P. (B) G.P. (C) H.P. (D) None [B] Coordinates of P, Q, R are (cos, sin), (cos, sin) and (cos, sin) respecively. Also, A  (–1, 0)  AP = (1  cos)2  sin 2  = 2 cos /2 AQ = (1  cos) 2  sin 2  = 2 cos /2 AR = (1  cos ) 2  sin 2  = 2 cos /2  AP, AQ, AP are in G.P. then    cos , cos , cos are also in G.P. 2 2 2  radius = tt–4 2 = 2 2 so, |2t – 4| = 4  2t – 4 = ± 4 i.e. t = 4, 0  centre = (4, 4); (0, 0)  satisfying x + y > 4  centre (4, 4), radius 2 2  equation of circle (x – 4)2 + (y – 4)2 = ( 2 2 )2  x2 + y2 – 8x – 8y + 16 + 16 = 8  x2 + y2 – 8x – 8y + 24 = 0 Q.5 Sol. The values of p for which the power of a point (2, 5) is negative with respect to circle x2 + y2 – 8x – 12y + p = 0 which neither touches the axis nor cuts them are (A) p (1, 2) (B) p (13, 27) (C) p (36, 47) (D) p (49, 52)  – 4C2 – 16C + 244 = 0  C2 + 4C – 61 = 0  C1 & C2 are roots of the equation  C1 + C2 = – 4 Q.7 For what number of real values of  line y x + + 1 = 0 touches circle  x2 + y2 – 2x – 2y + 1 = 0 (A) One (B) two (C) zero (D) four Sol. [C] Given circle is x2 + y2 – 2x – 2y + 1 = 0 [C] We have x2 + y2 – 8x + 12y + p = 0 Then centre and radius of the circle are (4,6) and 52 – p respectively.  circle neither cuts nor touches any one of the axes of co-ordinate of centre > radius then x coordinate of centre > radius i.e. 4 >  (x – 1)2 + (y – 1)2 = 1 Whose centre is (1, 1) and radius = 1 52 – p  p > 36 ...........(i) and y-co-ordinate of centre > radius 6> y + 1 = 0 touches the circle, then length  of perpendicular drawn from its centre to the line must be equal to its radius. If x + 52 – p  p > 16 ..........(ii) 1  1   =1 1 2   2  y  C(4,6) D(2,5) x O 1 1       + 1 = 2  2      D is interior point of the circle then CD < radius 5 < on squaring 52 – p 2 1 1 1        + 1 + 2     = 2 + 2         p < 47 .............(iii) From (i), (ii), (iii) we obtain 36 < p < 47   2 +  p  (36, 47) 1 +2+1+2 2  2  1    = 2 + 1    2    3 + 22 + 2 = 0 Q.6 C has two values C1 and C2 for which y = 2x + C touches a circle x2 + y2 – 4x –4y – 5 = 0 then C1 + C2 is equal to(A) 2 65 Sol. (B) 4 (C) – 4  22 + 3 + 2 = 0 its D < 0 since b2 – 4ac = 9 – 4 × 2 × 2 = – 7 < 0  There is no real value of  possible (D) – 2 65 [C] y = 2x + C touches the circle x2 + y2 – 4x – 4y – 5 = 0 x2 + (2x + C)2 – 4x – 4 (2x + C) – 5 = 0 Q.8 x + 4x + C + 4xC – 4x – 8x – 4C – 5 = 0 2 2 The length of the tangent drawn from any point on the circle x2 + y2 + 2gx + 2fy +  = 0 to the circle x2 + y2 + 2gx + 2fy +  = 0 is (A) – (B)  –  (C)  (D) 2  x2 + 4x2 + C2 + 4xC – 4x – 8x –4C – 5 = 0  5x2 + x (4C – 12) – (4C + 5 – C2) = 0 D=0 (4C – 12)2 – 4.5 (C2 – 4C – 5) = 0  16C2 + 144 – 96C – 20C2 + 80C + 100 = 0 Sol.  / [A] Let (x1, y1) is the point on the given circle x2 + y2 + 2gx + 2fy +  = 0  (x1,y1) satisfies the circle  x12 + y12 + 2gx1 + 2fy1 +  = 0  x12 + y12 + 2gx1 + 2fy1 = –  ......(i)  Length of tangent from (x1,y1) to the circle x2 + y2 + 2gx + 2fy +  = 0 is x12  y12  2gx1  2fy1   = – = (A) Sol. – Sol. If the chord of contact of the pair of tangents from P to the circle x2 + y2 = a2 which touches the circle x2 + y2 – 2ax = 0, then the locus of P is (A) y2 = a (a – 2x) (B) x2 + y2 = (x + a) 2 2 (C) x = a(a – 2y) (D) None of these 30º 30º  2 = 3 = g2  f 2 – c s1 5 a 2  b2 – 5 5 a 2  b2 – 5 1 5 =  a2 + b 2 – 5 = 5 3 3 5 20  a2 + b 2 = 5 +  a2 + b 2 = 3 3  (a, b) lies on circle x2 + y2 = r2 20  r2 = 3 r= 20 3 Q.12 Locus of mid point of chord AB of a circle x2 + y2 = 4 which subtends 90º angle at (1, 0) is(A) x2 + y2 + y + 3 = 0 (B) x2 + y2 – x – 3/2 = 0 (C) x2 + y2 – y – 3/2 = 0 (D) none of these Sol. [B] y A 75 300  5 3  1 = 2 10 3 1 = 2  1 : 2 = 1 : 2 Q.11 1 300   1 = 2  = 2  on squaring [A] Since (0, 0) lies on the circle 15x2 + 15y2 – 48x + 64y = 0  length of tangent from (0, 0) to first circle is 1 = 75 and lengths of tangent from (0,0) to the second circle 5 2 a 2  b2 – 5  tan 30º = =±a h2  k2  (h–a)2 = h2 + k2  locus of (h, k) is (x – a)2 = x2 + y2 or y2 = a (a – 2x) Sol. (D) (–a,–b) We know that tan [A] Let P(h, k) be the point of intersection of the tangents at the extrmities of the chord AB of the circle x2 + y2 = a2. since AB is the chord of contact of the tangents from P to this circle, its equation is hx + ky = a2. If this line touches the circle x2 + y2 – 2ax = 0, then The lengths of the tangents from any point on the circle 15x2 + 15y2 – 48x + 64y = 0 to the two circles 5x2 + 5y2 – 24x + 32y + 75 = 0 and 5x2 + 5y2 – 48x + 64y + 300 = 0 are in the ratio (A) 1 : 2 (B) 2 : 3 (C) 3 : 4 (D) None 20 3 60º h.a  k.0 – a 2 Q.10 (C) [C] (0,0)O Q.9 15 2 (B) 20 If circle x2 + y2 + 2ax + 2by + 5 = 0 subtends angle 60º at origin then (a, b) lies on circle whose radius is equal to - N (h,k) P O (1,0) x B Let N (h, k) be the middle point of chord AB which subtends 90º angle at (1, 0) since  APB = 90º  NA = NB = NP (since distance of the vertices from mid-point of the hypotenuse are equal) or (NA)2 = (NB)2 = (h – 1)2 + (k – 0)2 .......(i) 2  2   (x1, y1)   – a 1 , – a m1  lies on 2x+m2y+n2 = 0  n1   a 21   + m2  2  –  n1    But also BNO = 90º  (OB)2 = (ON)2 + (NB)2 –  – (NB)2 = (ON)2 – (OB)2 Q.13 Sol. The locus of a point such that the tangents drawn from it to the circle x2 + y2 – 6x – 8y = 0 are perpendicular to each other is (A) x2 + y2 – 6x – 8y – 25 = 0 (B) x2 + y2 + 6x – 8y – 5 = 0 (C) x2 + y2 – 6x + 8y – 5 = 0 (D) x2 + y2 – 6x – 8y + 25 = 0 Q.14 Sol. Two lines 1x + m1y + n1 = 0 and 2x + m2y + n2 = 0 are conjugate lines with respect to the circle x2 + y2 = a2 if (A) 12 + m1m2 = n1n2 (B) 12 + m1m2 + n1n2 = 0 (C) a2 (12 + m1m2) = n1n2 (D) 12 + m1m2 = a2 n1n2 [C] Since 1x + m1y + n1 = 0 and 2x + m2y + n2 = 0 are conjugate lines. let (x1,y1) be the polar then its polar is xx1 + yy1 = a  xx1 + yy1 – a = 0 2 2 a 2 12 a 2 m1m 2 – + n2 = 0 n1 n1 The Locus of the middle point of chords of the circle x2 + y2 = a2 which passes through the fixed point (h, k) is . (A) x2 + y2 – hx – ky = 0 (B) x2 + y2 + hx + ky = 0 (C) x2 + y2 – 2hx – 2ky = 0 (D) None of these Sol. [A] Let mid-point of circle is (x1, y1)  T = S1 xx1 + yy1 – a2 = x12 + y12 – a2  xx1 + yy1 = x12 + y12  x12 + y12 – xx1 – yy1 = 0 since it passes (h, k) 2  x1 + y12 –hx1 – ky1 = 0  locus of (x1, y1) is given by x2 + y2 – hx – ky = 0 Q.16 If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 – 2x + 8y – d = 0, then c + d = (A) 60 (B) 50 (C) 40 (D) 56 Sol. [B] S1  x2 + y2 + 4x + 22y + c = 0 S2  x2 + y2 – 2x + 8y – d = 0  S1 – S2 = 0  6x + 14 y + (c + d) = 0 since it passes the centre of S2 i.e. (1, – 4)  6 – 56 + (c + d) = 0  c + d = 50 Q.17 The equation of a circle is x2 + y2 = 4. The centre of the smallest circle touching this circle and the line x + y = 52 has the coordinates  7 7  3 3  , (A)  (B)  ,  2 2 2 2 2 2 and 1x + m1y + n1 = 0 – a2 x1 y = 1 = n1 1 m1  x1 = –  a 2 m1  –  + n2 = 0  n1   Q.15 [A] Given circle is (x – 3)2 + (y – 4)2 = 25 since locus of point of intersection of two perpendicular tangents is director circle, then its equation is (x – 3)2 + (y – 4)2 = 50  x2 + y2 – 6x – 8y – 25 = 0    – a212 – a2m1m2 + n1n2 = 0  – a2 (12 + m1m2) + n1n2) = 0  – a2 (12 + m1m2) = – n1n2  a2 (12 + m1m2) = n1n2  – [(h – 1)2 + (k – 0)2] = (h2 + k2) – 4 or 2(h2 + k2) – 2h + 1 – 4 = 0 or h2 + k2 – h – 3/2 = 0  locus is x2 + y2 – x – 3/2 = 0 n1  7 7   (D) None of these ,– (C)  –  2 2 2 2 – a 2 m12 a 2 1 ; y1 = n1 n1 Sol. [A] (A) 2 2 (0,0) Sol. O 2 B (B) 4 2 (C) 6 2 (D) 8 2 [B] (1,3) C 3 C D A A x+y=5 2 Here OB = radius = 2 and OC = 2 + =4 2 Q.20 The equation of a chord of the circle x2 + y2 + 4x – 6y = 0 is given by x + 2y = 0. The equation of the circle described on this chord as diameter is (A) 5(x2 + y2) + 28x + 14y = 0 (B) 5(x2 + y2) + 28x – 14y = 0 (C) x2 + y2 + 28x – 14y = 0 (D) 5(x2 + y2) – 28x – 14y = 0 Sol. [B] We have S + L = 0 (x2 + y2+ 4x – 6y)+  (x + 2y) = 0  x2 + y2+ x (4 + ) + y (2 – 6) = 0  4  , – ( – 3)  its centre is  – 2    4  , 3 – )   – 2   This centre must lie on x + 2y = 0 4 – + 2 (3 – ) = 0 2  –( + 4) + 4 (3 – ) = 0  –  – 4 + 12 – 4 = 0  – 5 = – 8   = 8/5  Reqd. circle is 8 (x2 + y2 + 4x – 6y) + (x + 2y) = 0 5 5x2 + 5y2 + 20x – 30y + 8x + 16 y = 0 5x2 + 5y2 + 28x – 14y = 0 5(x2 + y2) + 28x – 14y = 0 Q.21 The common chord of the circle x2 + y2 + 6x + 8y – 7 = 0 and a circle passing through the origin, and touching the line y = x, always passes through the point (A) (– 1/2, 1/2) (B) (1, 1) (C) (1/2, 1/2) (D) None of these The slope of OA = 1 = tan 1  cos = 2 , sin = 1 2  C = (0 + OC. cos, 0+ OC sin)  7 1 7 1   ,0  . =  0  . 2 2 2  2   7 7   , =  2 2 2 2 Q.18 The circle x2 + y2 – 2px = a2 – p2 lies within the circle x2 + y2 – 2qx = b2 – q2 if (A) p2 + q2 – a2 – b2 < 2 (pq – ab) (B) p2 + q2 – a2 – b2 > 2 (pq – ab) (C) p2 + q2 – a2 – b2 < 2 (pq + ab) (D) None of these Sol. [A] S1  x2 + y2 – 2px – a2 + p2 = 0 S2  x2 + y2 – 2qx – b2 + q2 = 0 C1 : (p, 0) r1 : p2  a 2 – p2 = a C2 : (q, 0) r2 : q 2  b2 – q 2 = b C1C2 = .........(i) .........(ii) (p – q) 2 = p – q r1– r2 = a – b = a – b since one circle lies within the other circle  C1C2 < |r1 – r2|  (C1C2)2 < |r1 – r2|2  (p – q)2 < (a – b)2  p2 + q2 – 2pq < a2 + b2 – 2ab  p2 + q2 – a2 – b2 < 2pq – 2ab  p2 + q2 – a2 – b2 < 2(pq – ab) Q.19 For the circle x2 + y2 – 2x – 6y + 1 = 0 the chord of minimum length and passing through (1, 2) is of length- B = 2.2 2 5–2 3 = 2 2 3 7 = 2 2 (1,2) Minimum length of chord = AB = 2AD The distance of (0, 0) from x + y = 5 2 is 5 The radius of the smallest circle = 2 2 1 Sol. [C] Let the second circle be x2 + y2 + 2gx + 2fy = 0 The common chord has the equation 2(g – 3)x + 2(f – 4) y + 7 = 0 But y =x touches the circle  x2 + x2+ 2gx + 2fx = 0 has equal roots i.e. f + g = 0  the equation of the common chord is 2(g – 3)x + 2(f – 4) y + 7 = 0 or (–6x – 8y + 7) + g (2x – 2y) = 0 which passes through the point of intersection of – 6x – 8y + 7 = 0 and 2x – 2y = 0  6x + 8y – 7 = 0 & x = y 6x + 8y – 7 = 0  14x = 7 7 1 x= = 2 14 1 1 x=y=  ,  2 2 Q.22 The members of a family of circles are given by the equation 2(x2 + y2) + x – (1 + 2) y – 10 = 0. The number of circles belonging to the family that are cut orthogonally by the fixed circle x2 + y2 + 4x + 6y + 3 = 0 is (A) 2 (B) 1 (C) 0 (D) None of these Sol. [A] 1  2  x +y + x– y–5=0 2 2 and x2 + y2 + 4x + 6y + 3 = 0 cut orthogonally; if 2 2. Q.23 The circle passing through three distinct points (1, t), (t, 1) and (t, t) passes through the point (A) (1, 1) (B) (–1, –1) (C) (–1, 1) (D) (1, –1) Sol. [A] Let the equation of circle is x2 + y2 + 2gx + 2fy + c = 0 It is passing through (1, t), (t, 1) and (t,t) than 1 + t2 + 2g + 2ft + c = 0 .....(i) 2 t + 1 + 2gt + 2f + c = 0 ......(ii) 2t2 + 2gt + 2ft + c = 0 ........(iii) on solving (i), (ii) & (iii), we get 2g (t – 1) + 2f (1 – t) = 0 or g – f = 0 and t2 – 1 + 2f (t – 1) = 0 f=– from (iii), 2t2 – t(t+1) – t (t+1)+ c = 0 [c = 2t]  x2 + y2 – (t + 1)x – (t + 1)y + 2t = 0  (x2 + y2 – x – y) – t (x + y – 2) = 0  P + Q = 0  P = 0 and Q = 0 2 2 Then x + y – x – y = 0 ..........(iv) x+y–2=0 ......(v) form (iv) & (v) x = 1 and y = 1  (1, 1) Q.24 If the radical axis of the circles x2 + y2 + 2gx + 2fy + c = 0 and 2x2 + 2y2 + 3x + 8y + 2c = 0 touches the circle x2 + y2 + 2x + 2y + 1 = 0, then (A) g = 3/4 and f  2 (B) g  3/4 and f = 2 (C) g = 3/4 or f = 2 (D) None of these Sol. [C] The given circles are S1 : x2 + y2 + 2gx + 2fy + c = 0 2  .2+2. 4 –  1  2  – .3=–5+3  4   3 (1 + 2) = – 2 2  2 – 3 – 32= – 4  32 – 2 – 1 = 0  32 – 3 +  – 1 = 0  3(– 1) + 1 ( – 1) = 0  ( – 1) (3 + 1) = 0 1  = 1, – 3  Two real values of  Hence There are two circles belonging to this family. ( t  1) =g 2 S2 : x2 + y2 + 3 x + 4y + c = 0 2 The equation of the radical axis of the two circles is S1 – S2 = 0 3  i.e.  2g –  x + (2f – 4)y = 0 2  i.e. (4g – 3)x + (4f – 8)y = 0 since it touches the circle x2+y2+2x + 2y + 1 = 0 S1 –S2 = 0 (4g – 3)(–1)  (4f – 8)(–1)  (4g – 3) 2  (4f – 8) 2 – =1 2, 2    (4g – 3)2 + (4f – 8)2 + 2(4g – 3) (4f – 8) = (4g – 3)2 + (4f – 8)2 Sol. The coordinates of the radical centre of the three circles x2 + y2 = 9, x2 + y2 – 2x – 2y = 5 and x2 + y2 + 4x + 6y = 19 are (A) (– 1, 1) (B) (1, – 1) (C) (1, 1) (D) (0, 0) [C] S1 = x2 + y2 – 9 = 0 S2 = x2 + y2 – 2x – 2y – 5 = 0 S3 = x2 + y2 + 4x + 6y – 19 = 0 S1 – S2 = 0 & S2 – S3 = 0  2x + 2y – 4 = 0  x + y – 2 = 0 ..............(1) from S2 – S3, – 6x – 8y + 14 = 0  3x + 4y – 7 = 0 ..........(ii) from (i) & (ii) we get 3x + 3y – 6 = 0 3x + 4y – 7 = 0 – – + –y +1=0 y=1 x=1  Radical centre is (1, 1) Q.26 Sol. The radical centre of three circles described on the three sides of a triangle as diameter is the (A) orthocentre (B) circumcentre (C) incentre (D) centroid [A] The radical centre of three circles described on the three sides of a triangle as diameter is the orthocentre of the triangle. Part-B Q.27 Sol. One or more than one correct answer type questions P(2, 2) is a point on the circle x2 + y2 = 4 and Q is another point on the circle such that arc 1 PQ = × circumference. The coordinates of 4 Q are (A) (–2, –2) (B) (2, –2) (C) (–2, 2) (D) None of these [B,C]   2 ,– 2   (4g – 3)(4f – 8) = 0  g = 3/4 or f = 2 Q.25 2, 2 Clearly from above diagram Required co-ordinates of Q are given by (– 2 , 2 ) and ( 2 ,– 2 ) Q.28  Sol. If A and B are two points on the circle x2 + y2 – 4x + 6y – 3 = 0 which are farthest and nearest respectively from the point (7, 2) then (A) A = (2 – 22, – 3 – 22)   (B) B = (2 + 22, – 3 + 22) (C) A = (2 + 22, – 3 + 22)   (D) B = (2 – 22, – 3 + 22) [A,B] r P 4 C 4 –3 – 2 –5 = =1 –5 2–7 If tan = 1,   = 45º Equation of PA is x–7 y–2 = =r 1/ 2 1/ 2 Slope of PC =  r r   lie on circle then ,2    7  2 2   r   7   2  2 2 +  r   2   – 2   r   4  7  2   r   – 3 = 0 + 6  2  2   r2 + 10 2 r + 34 = 0  r = –5 2 ± 4  –5 2 4 – 5 2  4  ,2   points are  7    2 2    (2 ± 2 2 , – 3 + 2 2 )  B (2 + 2 2 , –3 + 3 2 ) A (2 –2 2 , –3 –2 2 ) Q.29 Sol. If (a cos i, a sin i ) i = 1, 2, 3 represent the vertices of an equilateral triangle inscribed in a circle, then (A) cos 1 + cos 2 + cos 3 = 0 (B) sin 1 + sin 2 + sin 3 = 0 (C) tan 1 + tan 2 + tan 3 = 0 (D) cot 1 + cot 2 + cot 3 = 0 Q.31 Equations of a circle which touches the axes and x/a + y/b = 1, centre being in positive quadrant is x2 + y2 – 2rx – 2ry + r2 = 0, where r = (C) [A,B] (acos1, asin1) A a  b  a 2  b2 2 ab (A) Sol. a  b  a 2  b2 2 ab (B) (D) a  b  a 2  b2 a  b  a 2  b2 [B,C,D] (0,b) x y  1 a b B (acos2, asin2) r C (acos3, asin3) r ABC is an equilateral triangel Since (acosi; a sini) lie on the circle whose equation is x2 + y2 = r2 whose centre is (0, 0) and radius = r since centre of circle and centroid of ABC are the same point  centroid of ABC is given by a cos 1  a cos  2  a cos 3 =0 3 a sin 1  a sin 2  a sin 3 and =0 3  cos1 + cos2 + cos3 = 0 and sin1 + sin2 + sin3 = 0 Q.30 Sol.  (a,0) p=r r r  –1 a b  =r 1 1  a 2 b2  r r  –1 = r a b ab r   ±r  ab  ab r  ±r  ab  r= r = r= r = (,0) 1 a 2  1 b2 a 2  b2 =1 a 2b2 a 2  b2 =1 ab a 2  b 2 ] = ab  r [(a + b) ± [A,B] x2 + y2 + ax + by + c = 0 meets the x-axis  x2 + ax + c = 0 its roots are  &  +=a  = c  equation of whose roots are , &  are x2 – x( + ) +  = 0  x2 – x(–a) + c = 0  x2 + ax + c = 0  x2 + ax – a = 0 [ +  =   c = –a] and x2 – cx + c = 0 x y + =1 a b  circle touches If the circle x2 + y2 + ax + by + c = 0 meets the axis of x at (, 0) and (, 0) such that + = then (, ) are the roots of the equation. (A) x2 + ax – a = 0 (B) x2 – cx + c = 0 2 (C) x + bx – b = 0 (D) x2 – bx + b = 0 (,0) (r,r) ab a  b  a 2  b2 ab a  b  a 2  b2 &r= a  b  a 2  b2 (a  b ) 2 – (a 2  b 2 ) = ab a  b – a 2  b2 a  b  a 2  b2 2ab a  b – a 2  b2 a  b  a 2  b2 or 2 2 Q.32 The equation of one of the tangents from (1, 1) to a circle with its centre at (3, 0) is 3x + y – 4 = 0. The equation of the other tangent is(A) 5x – y – 4 = 0 (B) 3y – x – 2 = 0 (C) 3y + x – 4 = 0 (D) 3x – y – 2 = 0 Sol. [A,B,D] for (3, – 4)  9 + 16 + 6g – 8f + c = 0  6g – 8f + c + 25 = 0 .……(ii) for (– g, 0)  g2 – 2g2 + c = 0  g2 = c …….(iii) from (i) & (ii)  4g – 8f + 2c + 10 = 0 6g – 8f + c + 25 = 0 – + – – – 2g + c – 15 = 0  2g = c – 15 3x+y–4=0 (3,0) (1,1) mx – y + (1 – m) = 0 Let equation of tangent is y – 1 = m (n – 1) y – 1 = mx – m mx – y + (1 – m) = 0 3(3)  0 – 4 3m – 0  1(1 – m)  = 9 1 m2  1  5 = 2m  1 m2  1  25 (m + 1) = (2m + 1)2 .10  5m2 + 5 = 8m2 + 8m + 2  3m2 + 9m – m – 3 = 0  3m (m + 3) – 1 (m + 3) = 0  (m + 3) (3m – 1) = 0  m = – 3, 1/3 but m  –3 1 m= 3 x  1  – y + 1 –  = 0 3  3 x 2  –y+ =0 3 3  x – 3y + 2 = 0  3y – x – 2 = 0 10 (i) × 2 from (iii) c = g2 putting in this equation g2 – 2g – 15 = 0  g2 – 5g + 3g – 15 = 0  g(g – 5) + 3(g – 5) = 0  (g – 5) (g + 3) = 0  g = – 3, 5  C = 9, 25 when g = –3, 5 responsing when g = –3, c = 9, we get  2(–3) – 4f + 9 + 5 = 0 f=2 when g = 5, c = 25 we get 10 – 4f + 30 = 0 f = 10  Reqd. equation of circle is given by when g = – 3, f = 2, c = 9 x2 + y2 – 6x + 4y + 9 = 0 Ans. find when g = 5, f = 10, c = 25 x2 + y2 + 10x + 20y + 25 = 0 2 Q.33 Equation of circles which pass through the point (1, –2) and (3, –4) and touch the x-axis is(A) x2 + y2 + 6x + 2y + 9 = 0 (B) x2 + y2 + 10x + 20y + 25 = 0 (C) x2 + y2 – 6x + 4y + 9 = 0 (D) None of these Sol. Since required circles passes through (1, –2) and (3, – 4) and also touches x-axis Let equation of circle is x2 + y2 + 2gx + 2fy + c = 0 ………(1) Its centre is (– g, – f) Part-C Assertion-Reason type questions The following questions 34 to 37 consists of two statements each, printed as Assertion and Reason. While answering these questions you are to choose any one of the following four responses. (A) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (B) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion. (C) If Assertion is true but the Reason is false. (D) If Assertion is false but Reason is true (–g,–f) (–g,0) Q.34 Assertion (A) : Two distinct chords drawn from the point (3, 1) on the circle x2 + y2 – 3x – y = 0 are bisected by the x-axis. Reason (R) : If point of bisection is (h, 0) then equation of chord given by T = S1 passing through (3, 1) will be quadratic in h giving two distinct values of h. Sol. [A] x-axis  it touches x-axis at (–g, 0) so it is also passes (– g, 0) Now we have to find equation of circle which passes the three point (1, –2), (3, –4) and (–g, 0) for (1, – 2)  1 + 4 + 2g – 4f + c = 0  2g – 4f + c + 5 = 0 ……..(i) cuts the 3 circles orthogonally & is called as radical or orthogonal circle. Reason is correct but it is not correct explanation. P (h,0) A (3, 1) C Q.37 Assertion (A) : If a line L = 0 is tangent to the circle S = 0, then it will also be a tangent to the circle S + L = 0. Reason (R) : If a line touches a circle, then perpendicular distance of the line from the centre of the circle is equal to the radius of the circle. Sol. [B] (y = 0) S  x2 + y2 – 3x – y = 0 Let the mid point on the chord be (h, 0), hR then, equation of chord is given by L=0 T = S1 i.e., hx + 0.y – 3 1 (x + h) – (y + 0) 2 2 as shown in figure 1 = h2 – 3h i.e., hx – 3 3 y x+ h– – h2 = 0 2 2 2  it satisfies (3, 1),   3h – 3 3 1  h  – h2 = 0 2 2 2 h2 + 9 h–2=0 2 2 2h + 9h – 4 = 0 D = 92 – 4.2.(– 4) = 81 + 32 = 113 > 0 it shows that two such chords are possible Q.35 Sol. Q.36 Assertion (A) : Angle between line x + y = 3 and circle x2 + y2 – 2x – 4y – c2 = 0 will not depend on c. Reason (R) : As line passes through centre of circle so angle is 90º. [A]  the centre (1, 2) lies on the line x + y = 3 this is a diameter of the circle. Therefore, it  subtends angle at any point on the circle. 2 Assertion (A) : If three circles which are such that their centres are non-colliner, then exactly one circle exists which cuts the three circles orthogonally. Reason (R) : Radical axis for two intersecting circles is the common chord. Sol.[B] Assertion (A) If 3 circles which are such that their centers are non-collinear, then exactly one circle exists which Part-D Column Matching type questions Q.38 Column 1 (A) Circles x2 + y2 – 5x – 3 = 0 and 3x2 + 3y2+2x + 4y – 6 = 0 (B) x2 + y2 – 2x – 4y – 5 = 0, x2 + y2 – 4x – 2y – 3 = 0 (C) x2 + y2 = 9, x2 + y2 – 2x – 2y – 5 = 0 (D) Equation of common chord of circles x2 + y2 + 2x + 2y + 1 = 0 and x2 + y2 + 4x + 3y + 2 = 0 is Sol.  Column 2 (P) Line joining centres of two circles is x + y = 3 (Q) Radical axis of this two circles is 17x + 4y + 3 = 0 (R) Radical axis of two circles x + y = 2 (S) 2x + y + 1 = 0 A Q, B P, C  R, D  S (A) S1 : x2 + y2 – 5x – 3 = 0 ........(i) S2 : 3x2 + 3y2 + 2x + 4y – 6 = 0 2 4  x2 + y2 + x + y – 2 = 0 ....(ii) 3 3  S1 – S2 = 0 2  4    – 5x – x  – y – 3 + 2 = 0 3 3    (– 15x – 2x) – 4y – 3 = 0  17x + 4y + 3 = 0 (B) S1 : x2 + y2 – 2x – 4y – 5 = 0 S2 : x2 + y2 – 4x – 2y – 3 = 0 S1 – S2 = 0  – 2x – 2y – 2 = 0 x+y+1=0 C1 : (1, 2) C2 (2, 1) y – 2 = – 1 (x – 1) y – 2 = –x + 1 x+y=3  line joining centre of two circle is x + y = 3 (C) S1 : x2 + y2 = 9 .........(i) S2 : x2 + y2 – 2x – 2y – 5 = 0 ........(ii) S1 – S2 = 0  2x + 2y – 4 = 0 x+y=2 (D) S1 – S2 = 0  2x + y + 1 = 0 Q.39 Sol. Column-1 Column-2 (A) Number of values of a for which (P) 4 the common chord of the circles x2 + y2 = 8 and (x – a)2 + y2 = 8 subtends a right angle at the origin is (Q) 2 (B) A chord of the circle (x–1)2 + y2= 4 lies along the line y = 22 3 (x–1). The length of the chord is equal to (C) The number of circles touching all (R) 0 the three lines 3x + 7y = 2, 21x + 49y=5 and 9x + 21y=1 are (D) If radii of the smallest and largest (S) 1 circle passing through the point ( 3 , 2 ) and touching the circle (D) (0, 2 ) 2 3 ( 3 , 2 ) lies inside the circle S = x2 + y2 – 2 2 y – 2 = 0 r1 = 2  3 and r2 = 2  3 r r  mean = 1 2 = 2 2 Q.40 x2 + y2 – 2 2 y – 2 = 0 are r1 and r2 respectively, then the mean of r1 and r2 is A Q, B P, C  R, D S (A) equation of common chord is a x= 2 Since it subtend  2x  x2 + y2 =    a   at the origin, 2 2 4 ) x2 + y2 = 0 a2  co-eff of x2 + co-eff of y2 = 0 i.e., (1 –  1– 4 a2 +1=0 a=± 2 (B) length of chord = 2 r 2  p 2 p=0 (1,0) r=2 p  length of the chord = 2 × 2 = 4 (C) Since all three lines are parallel, a circle touching all the lines. r=2 Sol. Column-1 Column-2 (A) Number of common tangents of (P) 1 the circles x2 + y2 – 2x = 0 and x2 + y2 + 6x – 6y + 2 = 0 is (B) Number of indirect common (Q) 2 tangents of the circles x2 + y2 –10y – 4x + 4 = 0 & x2 + y2 – 6x – 12y –55 = 0 is (C) Number of common tangents of (R) 3 the circles x2 + y2 – 2x – 4y = 0 & x2 + y2 – 8y – 4 =0 is (D) Number of direct common tangents (S) 0 of the circles x2+y2+2x–8y + 13 = 0 & x2 + y2 – 6x – 2y + 6 = 0 is A R, B S, C  P, D Q (A) C1  (1, 0), r1 = 1 C2 = (–3, 3), r2 = 4 C1C2 = r1 + r2 = 5 # tangents = 3 (B) C1 = (5, 2), r1 = 2 C2 = (3, 4), r2 = 10 C1C2 < r2 – r1 # indirect tangents = 0 (C) C1  (1, 2), r1 = 5 C2  (0, 4), r2 = 2 5 C1C2 = 5 = r2 – r1  C1 touches C2 internally  # tangents = 1 (D) C1  (–1, 4), r1 = 2 C2  (3, 1), r2 = 2 C1C2 > r1 + r2  # direct tangents = 2 EXERCISE # 2 (PART-A) Q.No. Ans. Q.No. Ans. 1 B 2 B 3 B 4 A 5 C 6 C 7 C 8 A 9 A 10 A 11 C 12 B 13 A 14 C 15 16 19 B 20 B 21 C 22 A 23 A 25 26 B 18 A 24 A 17 A C C A (PART-B) Qus. 27 28 29 Ans. B,C A,B A,B 30 31 A, B A,B,C,D 32 33 B B,C (PART-C) Qus. 34 35 36 37 Ans. A A B B (PART-D) 38. A Q, B P, C  R, D S 40. A R, B S, C  P, D Q 39. A Q, B P, C  R, D S