Circle Exercise # 5

EXERCISE # 5 Q.1 An isosceles right angled triangle whose sides are 1, 1,  2 lies entirely in the first quadrant with the ends of the hypotenuse on the coordinate axes. If it slides prove that the locus 32 of its centroid is (3x – y)2 + (x – 3y) 2 = . 9 (3x – y)2 + (x – 3y)2 = Q.2 Sol. (0, 2 sin) B (3h,– 2 cos,3k– 2 sin) 1  Sol. (h,k)  2) 1 ( 45°  A ( 2 cos,0) 32 9 In a right-angled triangle, the length of the sides are a and b(0 < a < b). Show that the radius of the circle passing through the midpoint of the smaller side and touching the hypotenuse at its mid-point is b C 45– Equating cos  and sin  from (1) & (2) and using cos2 + sin2 = 1 gives a 2  b 2 /4a. We have to choose the perpendicular sides along x axes of co-ordinates so that the hypotenuse is a y a b + = 1, which is a tangent at mid-point  ,  b 2 2 B (O,b) b P(a/2,b/2) 1 ( ( 2) 45°  (cos,0) Slope of AC = tan(180 – ( + 45)) = tan(135 – ) = C  ( 2 cos + cos(135 – ), 0 + sin(135 – ) = (3h – 2 cos , 3k – 2 sin ) comparing y – co-ordinat gives sin (135 – ) = 3k – 2 sin    1 (cos  + sin ) = 3k – 2 sin  2   cos  + 3 sin  = 3 2 k ……(1) slope of BC = tan (45 – ) C  (0 + 1.cos (45 – ), 2 sin + 1.sin(45 – )) = (3h – 2 cos , 3k – 2 sin ) comparing x-co-ordinate gives cos (45 – ) = 3h – 2 cos    1 2 (cos  + sin ) = 3h – 2 cos  A(a/0) Q (a/2,0) O 2) The euation of the cricle is given by 2 2 a b x y     x –  +  y –  +    – 1 = 0 2 2 a b    a  it passes through mid-point Q of OA i.e.  ,0  2   b2 b2  – = 0 or  = 4 2 2 Reuquird circle is 2 2 b2  x y  a b     – 1 = 0 x –  + y –  + 2 a b  2 2    b b 2   or x2 + y2 + x  – a  + y – b  +   2 2a    a 2  b2 b2 – =0 4 4 b 2 – 2a 2 a 2 – b2 b x– y+ =0 2a 4 2  r2 = g2 + f2 – c or x2 + y2 + 2 2 2 2  b 2 – 2a 2   +  b  – (a – b )  r2 =   4a  4 4   a2 b2 b2 a2 b2 = + – + – + 4 4 16 4 4 16a 2 x2 + y2 – 4x – 6y + 9 = 0 x2 + y2 – 20x – 22y + 121 = 0 b4 = b4  a 2b2 16a r= or Q.4 2 b4  a 2b2 16a 2 = b 4a a 2  b2 Sol. Q.3 Find the equation of the circle passing through the point A(4, 3) & B(2, 5) & touching the axis of y. Also find the point P on the y-axis such that the angle APB has largest magnitude. The locus of the mid-point of a chord of the circle x2 + y2 = 4 which subtends a right angle at the origin is (A) x + y = 2 (B) x2 + y2 = 1 (C) x2 + y2 = 2 (D) x + y = 1 [C] As we have to find locus of mid-point of chord and we know perpendicular from centre bisects the chord A (–g,–f) O Sol. (0,–f) (C) 2 B 45º A C(h,k) S clearly OAC = 45º Let equation of circle is x2 + y2 + 2gx + 2fy + c = 0 since circle passes through A(4, 3), B(2, 5), C(0, –f) passes (4, 3)  16 + 9 + 8g + 6f + c = 0  8g + 6f + c + 25 = 0 …..(i) passes (2, 5)  4 + 25 + 4g + 10f + c = 0  4g + 10f + c + 29 = 0 ….(ii) passes (0,–f)  f2 – 2f2 + c = 0 2 f =c ……(iii) from (i), (ii) we get or  14f + c + 33 = 0  f2 + 14f + 33 = 0  f2 + 11f + 3f + 33 = 0  f(f + 11) + 3(f + 11) = 0  (f + 11) (f + 3) = 0 f = – 3, – 11  c = 9, 121 respectively from (i) when f = –3, c = 9 then 8g + 6(–3) + 9 + 25 = 0  8g – 18 + 34 = 0  8g = –16 g=–2 and when f = –11, c = 121, then 8g + 6(–11) + 121 + 25 = 0  8g – 66 + 121 + 25 = 0  8g + 146 – 66 = 0  8g = – 80 g = – 10  Reqd. circle is B S OC = sin 45º OA  OC = 2 2 = 2 h 2  k 2 = OC  h2 + k2 = OC2 or x2 + y2 = 2 whcih is the required equation of locus of mid-point of chord subtentding right angle at centre. 8g + 6f + c + 25 = 0 8g + 25f + 2c + 58 = 0 – – – – –14f – c – 33 = 0 2 Q.5 From the origin chords are drawn to the circle (x –1)2 + y2 = 1. The equation of the locus of the mid-points of these chords is.......... Sol. For the equation of circle x2 + y2 –2x = 0, Let the mid-point of chords be (h, k)  Equation of chord bisected at the point is T = S1  h2 + k2 – 2h = xh + yk – (x + h) Which passes through (0, 0)  h2 + k2 – 2h = – h  h2 + k2 – h = 0  Locus is x2 + y2 – x = 0 Q.6 From the point A (0, 3) on the circle x2 + 4x + (y –3)2 = 0, a chord AB is drawn and extended to a point M such that AM = 2AB. The equation of the locus of M is ................... Sol. C2 A 5 (x + 2)2 + (y – 3)2 = 4 let if M (h, k) where B is mid-point of A and M.  h k 3 B  ,  2 2  But AB is the chord of circle x2+4x+(y–3)2 = 0 Thus B must satisfy above equation 4 (5,2) 3 O C1 4 B 5x–12y=10 2 h2 4h 1  + +  (k  3) – 3 = 0 4 2 2  2 2  h + k + 8h– 6k + 9 = 0  locus of M is the circle x2 + y2 + 8x – 6y + 9 = 0 i.e. Q.7 Sol. Lines 5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 touch a circle C1 of diameter 6. If the centre of C1 lies in the first quadrant, find the equation of the circle C2 which is concentric with C1 and cuts intercepts of length 8 on these lines. Since, 5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 are both perpendicular tangents to the circle, C1  OABC forms a square. Let the centre co-ordinate be (h, k) where OC = 6, OA = 6 and OB = 6 2 6 O(h,k) 6 5x – 12y = 10 C To obtain euqtion of circle concentric with C1 and making on intercept of lenght 8cm 5x + 12y = 10 and 5x – 12y = 40  required equation of circle C2 has centre (5, 2) and radius 5.  (x – 5)2 + (y – 2)2 = 52  x2 + y2 –10x – 4y + 4 = 0 Q.8 Let a given line L1 intersect the x and y axes at P and Q respectively. Let another line L2, perpendicular to L1, cut the x and y axes at R and S, respectively. Show that the locus of the point of intersection of the lines PS and QR is a circle passing through the origin. Sol. Let the equation of L1 be x cos + ysin = P1 Then any line perpendicular to L1 is given by y L2 Q S 5x – 12y = 40 A R O L1 B  5, – 15   12  5h  12k – 10 =3 13  5h+12k–10 = ± 39 and 5h – 12k – 40 = ± 39 on solving above equations. The co-ordinates which lie in Ist quadrant are (5, 2)  Centre for C1(5, 2)  x P x sin – y sin = p2 Where p2 is a variable. Then L1 meets x-axis at P(p1sec, 0) and y-axis at Q (0, p, cosec) Similarly L2 meets x-axis at R(p2 cosec, 0) and y-axis at S(O, – p2 sec) Now equation of PS is – x x + =1 p 1 sec  – p 2 sec   x x – = sec  p1 p2 ...........(i) Smilarly equation of QR is x x + =1 P 2 cosec  P1 cosec   Q.10 x y + = sec P2 P1 ...........(ii) Locus of point of intersection of PS and QR can be obtained by eliminating variable P2 from (i) and (ii)  Q.9  x  x y i.e.  + = cosec  – sec    p1  y P1    (x – p1sec ) x + y2 = p1 y cosece   x2 + y2 – p1 x seca – p1y cosec = 0 Which is a circle through origin. Sol. The equations of the tangents drawn from the origin to the circle x2 + y2 –2rx –2hy + h2 = 0, are (A) x = 0 (B) y = 0 (C) (h2 – r2)x –2rhy = 0 (D) (h2 – r2)x + 2rhy = 0 [C] Since tangents are drawn from origin. so let equation of tangent be y = mx where m is slope of tangent. A The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is x + y – xy + k (x2 + y2)1/2 = 0. Find k. (–r,h) B Q(0,q) Length of perpendicular from origin = radius  (h,k) (2,2) Sol. r (0,0) O mr – h m2  1  m2r2 + h2 – 2mrh = r2 (m2 + 1) 2 O =r m= P(p,o) – r2  h2 h2 – r2 = 2rh 2hr x y Let the equation of third side be + =1 p q or h2 + m2r2 – 2mhr = r2 + m2r2 or 0.m2 + 2mhr + (r2 – h2) = 0 Since OPQ is an Right angled triangle, whose circumcentre will bethe mid-point of PQ. Let (h, k) is the circum centre of the triangle. Since line PQ touches the given circle  Perpendicular distance from centre of circle = radius of circle Centre is (2, 2) and given line is x y + = 1 i.e qx + py – pq = 0 p q  m = , m = 2q  2p – pq q 2  p2 2(2k) + 2(2h) – 4hk =  2 4h 2  4k 2  h + k – hk  h 2  k 2 = 0  locus is x + y – xy ± k=1 Tangents are x = 0 for m =  and (h2–r2)x – 2rhy = 0 Q.11 If a circle passes through the point (a, b) and cuts the circle x2 + y2 = k2 orthogonally, then the equation of the locus of its centre is (A) 2ax + 2by –(a2 + b2 + k2) = 0 (B) 2ax + 2by –(a2 – b2 + k2) = 0 (C) x2 + y2 –3ax – 4by + (a2 + b2 – k2) = 0 (D) x2 + y2 –2ax – 3by + (a2 – b2 – k2) = 0 Sol. [A] Let x2 + y2 + 2gx + 2fy + c = 0 cuts x2 + y2 = k2 orthogonally, if 2g1g2 + 2f1f2 = c1 + c2 =2 since (h, k) is mid-point of PQ  P = 2h, q = 2k x 2  y2 = 0 h2 – r2 2hr  –2g.0 – 2f.0 = c – k2  c = k2 ...................(i) 2 2 Also, x + y + 2gx + 2fy + c = 0 passes through (a, b)  a2 + b2 + 2ga + 2fb + c = 0 ......(ii)  Required equation of locus of centre is – 2ax – 2by + a2 + b2 + k2 = 0  2ax + 2by – (a2 + b2 + k2) = 0 Q.12 Sol. 2 B If the circle C1 : x + y = 16 intersects another circle C2 of radius 5 in such a manner that common chord is of maximum length and has a slope equal to 3/4, then the coordinates of the centre of C2 are......... P(h,k) A C1 : x2 + y2 = 16 C2 : (x – h)2 + (y – k)2 = 25 common chords by S1 – S2 = 0 is 2hx + 2ky = h2 + k2 – 9 O The equation of chord AB is h y – k = – (x – h) k  hx + ky = h2 + k2 the combined equation of OA and OB is homogeneous equation of second degree obtained by the help of the given circle and the chord AB and is given by h 3 = (given) k 4  its slope = – If p be the length of perpendicular on it from the centre (0, 0) of C1 of radius 4, then p= h2  k2 – 9  hx  ky   hx  ky  x2+y2+(2gx + 2fy)  2  +c   =0  h  k2   h2  k2   2(h2 + k2) + 2 (gh + fk) + c = 0 c  h2 + k2 + gh + fk + = 0 2  Required eqution of locus is c x2+ y2 + gx + fy + =0 2 4h 2  4k 2 Also, the lenght of the chord is 2 r 2 – p2 = 2 42 – p 2 The chord will be of maximum length, if  = 0 or h2 + k2 – 9 = 0 16 2 h =9 9 or h = ± 9/5 or h2 + k=  Q.14 If the two circles (x –1)2 + (y –3)2 = r2 and x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then (A) 2 < r < 8 (B) r < 2 (C) r = 2 (D) r > 2 Sol. [A] As, if the two circles intersect in two distinct point 12 5  9 – 12   centres are  ,  5 5   9 12  and  – ,   5 5 Q.13 Sol. Let S  x + y + 2gx + 2fy + c = 0 be a given circle. Find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin. 2 x2 + y2 + 2gx + 2fy + c = 0 2 distance between |r1 – r2| and |r1 + r2| 2 Let P(h, k) be the foot of perpendicular drawn from origin, 0(0, 0) on the chord AB of the given circle such that the chord AB subtends a right angle at the origin.  i.e. |r – 3| < centres lies between (4 – 1) 2  (1  3) 2 < |r + 3|  |r – 3| < 5 < |r + 3|  r < 8 or r > 2 20   2  8    A a2 b2 – >0 4 2  a2 > 2b2 Q.21 Sol. = Q.22 Sol. D B Again, Let S is a circle with centre at C and AB is given chord and AD subtent angle 2/3 at the centre and D be the mid-point of AB and let its coordinates are (h, k) Now DCA = [D] Let (h, k) be the centre of the circle which touches the circle x2 + y2 – 6x – 6y + 14 = 0 and y-axis. Now, x2 + y2 + 2(–3)x + 2(–3)y + 14 = 0, the centre of this circle is (3, 3) and radius is C S The locus of the centre of a circle, which touches externally the circle x2 + y2 – 6x – 6y + 14 = 0 and also touches the y-axis, is given by the equation: (A) x2 – 6x – 10y + 14 = 0 (B) x2 –10x – 6y + 14 = 0 (C) y2 – 6x – 10y + 14 = 0 (D) y2 – 10x – 6y + 14 = 0 Again 1 1 2  (BCA) = . = 2 2 3 3 DA CA =   sin sin 3 2  DA = CA sin/3 32  32 – 14 = 18 – 14 = 2 3 3 3 3 . = 4 2 2 since the circle touches y-axis, the distance from its centre to y-axis must be equal to its radius, therefore its radius is h. Again the circles meets externally therefore the distance between two centres = sum of the radii of the two circles. Hence (h – 3)2 + (k – 3)2 = (2 + h)2 i.e. h2 + 9 – 6h + k2 + 9 – 6k = 4 + h2 + 4h i.e.k2 – 10h – 6k + 14 = 0 Thus the locus of (h, k) is y2 – 10x – 6y + 14 = 0 Now, in ACD, CD2 = CA2 – AD2 The equation of the locus of the mid-points of the chord of the circle 4x2 + 4y2 – 12x + 4y + 1 = 0 that subtend an angle of 2/3 at its centre is.....................  16x2 + 16y2 – 48x + 16y + 31 = 0 CD2 = 2 9 1  But CD2 = (h–3/2)2 +  k   = 16 2  on genralising, we get 2 Q.23 Consider a family of circles passing through two fixed points A (3, 7) and B (6, 5). Show that the chords in which the circle x2 + y2 – 4x – 6y – 3 = 0 cuts the members of the family are concurrent at a point. Find the coordinate of this point. Sol. The equation of the circle on the line joining the points A(3, 7) and B(6, 5) as diameter is (x – 3) (x – 6) + (y – 7) (y – 5) = 0 .............(i) and the equation of the line joining the points A(3, 7) and B(6, 5) is 1 =0 4  3 –1 its centre is  ,  and radius 2 2  2 2 1 3  – 1     – 4 2  2  = 3 9 1 1 =  – 2 4 4 4 2 9 3 1   x –  + y   = 16 2 2   Given circle is 4x2 + 4y2 –12x + 4y + 1 = 0 i.e. x2 + y2 – 3x + y + 9 27 9 – = 4 16 16 y–7= 7–5 (x – 3) 3–6  2x + 3y – 27 = 0 ..............(ii) Now the equation of family of circles passing through the point of intersection of (i) and (ii) is S + P = 0  (x – 3) (x – 6) + (y – 7) (y – 5) + (2x + 3y – 27) = 0  x2 – 6x – 3x + 18 + y2 – 5y – 7y + 35 + 2x + 3y – 27 = 0  S1  x + y – x(2 – 9) + y (3 – 12) + (53 – 27) = 0 ...............(iii) Again the circle, which cuts the memners of family of circles, is 2 2  2x – 5x + 3y – 6y + 56 – 27 = 0  13sin 2  – 9 sin 2  = 4 sin 2  = 2 sin  (h  2) 2  (k – 3) 2 = 2 Q.25 (a) The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Find the equation of the circle with AB as a diametre. (b) Find the the intervals of values of a for which the line y + x = 0 bisects two chords drawn from a point to  1  2a 1  2a    the circle ,  2  2   2x2 + 2y2 – (1+ 2 a) x – (1 – 2 a) y = 0. Sol. Equation of any circle passing through the point of intersection of x2 + y2 – 2x = 0 and y = x is  x2 + y2 – (2 + )x + y = 0 [B,D] Centre of the circle x2 + y2 + 4x – 6y + 9 sin2 + 13 cos2 = 0 is C(–2, 3) and its radius is given by A P (h,k) = (x2 + y2 – 2x) + (y – x) = 0 (A) x2 + y2 + 4x – 6y + 4 = 0 (B) x2 + y2 + 4x – 6y – 9 = 0 (C) x2 + y2 + 4x – 6y – 4 = 0 (D) x2 + y2 + 4x – 6y + 9 = 0 Sol. 13(1 – cos2 ) – 9 sin 2   (h + 2)2 + (k – 3)2 = 4  h2 + 4 + 4h + k2 + 9 – 6k = 4  h2 + k2 + 4h – 6 k + 9 = 0 Thus, required equation of the locus is x2 + y2 + 4x – 6y + 9 = 0  x{2–9+4}+y (3–12+6) + (53–27+3) =0 The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x – 6y + 9 sin2  + 13 cos2  = 0 is 2. The equation of the locus of the point P is- =   {x(2– 9) + y(3– 12) + (53 – 27)} – {– 4 x – 6y – 3} = 0 Q.24 13 – 13 cos2  – 9 sin 2  Let (h, k) be any point P and APC = , PAC = /2 i.e. triangle APC is a right angle triangle. AC 2 sin  Thus, sin = = PC (h  2) 2  (k – 3) 2 S2  x2 + y2 – 4x – 6y – 3 = 0 ..............(iv) and the equation of common chord to circles S1 and S2 is – S1 – S2 = 0  (5x – 6y + 56) +  (2x + 3y – 27) = 0 Which represents equatious of two straight lines passing through the fixed point whose coordinates are obtaind by solving the two equations. 5x + 6y – 56 = 0 and 2x + 3y – 27 = 0 on solving, weget x = 2 and y = 23/3 = C (–2,3) 2 – , its centre is   2   2  For AB to be the diameter of the requried circle the must be on AB. i.e. 2 +  = –   2 = – 2   = – 1 Therefore, equation of the required circle is x2 + y2 – (2 – 1)x – 1.y = 0  x2 + y2 – x – y = 0  x2 – x + y2 – y = 0 B 2 (–2)  (3) – 9 sin  – 13 cos  2 = 2 2 4  9 – 9 sin 2  – 13 cos2  2 2 1 1 1    x –  + y –  = 2 2 2     (b) 2x2 + 2y2 – (1 + 2 a)x – (1– 2 a)y = 0  1  2a  x –  x2 + y2 –   2     1 – 2a   y = 0  2    (6 2 a)2 – 4(8)(1+2a2) > 0  72a2 – 32 (1 + 2a2) > 0  72a2 – 32 – 64a2 > 0  8a2 – 327 > 0  a2 – 47 > 0  a2 > 4 a<–2a>2  a  (–, –2)  (2, ) since, y + x = 0 bisects two chords of this circle, mid-point of the chords must be of the form (– ). y y+x=0 O x –  1  2a 1  2a    ,  2 2   Q.26 For each natural number k, let Ck denotes the circle with radius k centimetres and centre at the origin. On the circle Ck, -particle moves k centimetres in the counter-clockwise direction. After completing its motion on Ck, the particle moves to Ck+1in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0). If the particle crosses the positive direction of the x-axis for the first time on the circle Cn then n = .............. Sol. It is given that C1 has centre (0, 0) and radius equal to 1 similarly; C2 has centre (0, 0) and radius 2, and Ck has the centre (0, 0) and radius k.  Equation of the chord having (,–) as mid-point is T = S1  1  2a   (x + ) – x + y (– ) –   4     1  2a   – = 2 +(–)2 –   4     1 – 2a    (y – )  4     1 – 2a    (– )  4    A4 A3  4x – 4y – (1+ 2 a)x – (1– 2 a) y A6 A5 – (1– 2 a)y + (1+ 2 a) = 42 + 42 –(1 +  4x + 4y –(1 + 2 a) x. – (1– 2 a)y A1 the  1 – 2a  (1  2a )(1  2a )  1  2a  –  –4  4   2   2  2     – (1 – 2a )(1 – 2a ) 2 = 8a2 – 2 2 a  O A0 (1,0) C1 2 a) . 2 + (1– 2 a) . 2 = 8a2 – (1 + 2 a) + (1 – 2 a) But this chord will pass through  1  2a 1 – 2a   point  ,  2  2   A1 A2 [(1+ 2 a – 1 + 2 a)] = 82– 2 2 a 1  4 2 a – [2 + 2 ( 2a ) 2 ] = 82– 2 2 a 2 [ (a + b2)+(a–b)2 = 2(a2 + b2)]  82 –6 2 a + 1 + 2a2 = 0 But this quadratic equation will have two distinct roots if C2 C3 C4 Now partical starts it motion from (1, 0) and moves 1 radian on first circle then particle shifts form C1 to C2 After that particle moves 1 radian on C2 and than particle shifts from C2 to C3 similarly, particle move on n circles Now n  2 because particle crosses the x-axis for the first time on Cn, then n is least positive integer. Therefore. n = 7 is the answer. Consider a curve ax2 + 2hxy + by2 = 1 and a point P not on the curve. A line drawn from the point P intersects the curve at points Q and R. If the products PQ. PR is independent of the slope of the line, then show that the curve is a circle (b) Two vertices of an equilateral triangle are (–1,0) and (1,0) and its third vertex lies above the x-axis. Find the equation of its circumcircle. - Q.27 (a) 8 Sol. (a) The given circle is ax2 + 2hxy + by2 = 1 Let the point P not lying on (i) be (x1,y1), Let  be the indintion of the line through P whin intersects the given curve at Q and R. Then equation of line through Pis x – x1 y – y1 = =r cos  sin   x = x1 + r cos, y = y1 + r sinq for point Q and R, above point must lie on (i)  a (x1 + rcos)2 + 2h (x1+rcos (y1 + r sin ) + b(y1 + rsin)2 = 1  (acos2 + 2hsincos + b sin2)r2 + 2(ax1cos + hx1sin + hy1cos+by1sin)r2 +2 (ax1 cos + hx1sin + hy1cos + by1sin) r + (ax12 + 2x1y1 + by12 – 1) = 0 It is quadratic in r, giving two values fo r as PQ and PR. Q PQ.PR =  k its centre O will coincide with centroed G  0,   3  1  2 2k  and radius r = OA = i.e.  0, = 3 3 3  2   2  1   =   circle is x2 +  y – 3   3 2y  x2 + y2 – –1=0 3 Q.28 The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to the circle x2 + y2 = 1 pass through the point........... Sol. A point on the line 2x + y = 4 is of the form (h, 4 – 2h) Equation of the chord of contact is T = 0 i.e. hx + (4 – 2h)y = 1 or (4y – 1) + h(x – 2y) = 0 This line passes through the point of intersection of 4y –1 = 0 and x – 2y = 0 1 1 i.e. through the point  ,  2 4 Q.29 Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C. (A rational point is a point both of whose coordinates are rational numbers) Sol. Equations of any circle C with centre at ax12  2hx1y1  by12 – 1 a cos2   2h sin  cos   b sin 2  Here, ax1 + 2hx1y1 + by12 – 1  0 cos (x1,y1) does not lie on (i),Also, a cos2 + 2h sincos + bsin2 = a + 2h sincos + (b – a)sin2 = a + sin (2hcos + (b – a)sin) 2 = a +sin 4h 2  (b – a ) 2 . [cossin+sincos] where tan = =a+ b–a 2h (0, 4h 2  (b – a ) 2 sin sin ( + ) which will be independent of , if 4h2 + (b – a)2 = 0  h = 0 and b = a  equation (i) reduees to 1 x2 + y2 = a which is a circle. 2 2 ) is given by (x – 0)2 + (y – 2 )2 = r2 or x2 + y2 –2 2 y + 2 = r2 ..........(i) where r > 0. Let (x1,y1), (x2, y2), (x3, y3) be three distinct rational point on(i). since a straight line parallel to x-axis meets a circle in atmost two points, either, y1,y2 or y1,y3 putting these in (i), we get x12  y12 – 2 2 y1 = r2 – 2 ..........(ii) x 22  y 22 – 2 2 y 2 = r2 – 2 ..........(iii) A (0,k)(0, 3 ) (b) 2 2k/3 x 32  y32 – 2 2 y3 = r2 – 2 .........(iv) subtracting (ii) from (iii), we obtain 2 p1 – 2 q1 = 0 O (0,k/3) B C D (0,0) (1,0) (–1,0) here BC = 2 = AB = AC as the traingle is euilateral. its vertex will be lie on the right bisector of BC. i.e. in y-axis. let it be (0, k) where k is +ve and equal to 2sin 60º = 3 where p1 = x 22 + y 22 – x12 – y12 q1 = y2 – y1 Subtreting (ii) form (iv), we obtain p2 – 2 q2 = 0 where p2 = y3 – y2 x32 + y32– x12 – y12 Now p1, p2, q1, q2 are rational numbers. Also either q1  0 or q2  0. If q1  0. then 2 =p2/p2 In aby case This is a contradiction. Q.30 = Sol.   C12 1  [r{2cos cos } + 2rcos ] 3 3 1 = [rcos + 2r cos] = r cos 3 1     and q = r sin   –   r sin      2r sin  3  3 3    7 =  1        r sin   –   sin      2r sin  3    3 3    =      1   –  – ––   3 3 3 3   2r sin  . cos 3 r 2 sin 2 2  (B) 1 (C) 3 (D) 4 S1  x2 + y2 – 4 = 0 C1  (0, 0) ; r1 = 2 S2  x2 + y2 – 6x – 8y – 24 = 0 C2  (3, 4) ; r2 = 7  C1C2 = 5  C1C2 = r2 – r1 So, only one tangent possible Q.31   = [B] C2      1   –  – ––   3 3 3 3   2r cos  . cos 3 r 2 cos 2 2     The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 – 6x – 8y = 24 is (A) 0 Sol. 2 is a rational number.      1  [r{2sincos } + 2rsin ] 3 3 1 = [r{sin} + 2r sin] = r sin 3 Now, (p, q)= (rcos, rsin) lies on x2 + y2 = r2 which is called C1. = C1 and C2 are two concentric circles, the radius of C2 being twice that of C1. From a point P on C2, tangents PA and PB are drawn to C1. Prove that the centroid of the triangle PAB lies on C1. Let the point P be (2r cos, 2r sin) we have OA = r, OP = 2r As OAP is a right angled triangle. C1 C2 A –  P    Q.32 If two distinct chords drawn from the point (p,q) on the circle x2 + y2 = px + qy (where pq  0) are bisected by the x-axis then (A) p2 = q2 (B) p2 = 8q2 2 2 (C) p < 8q (D) p2 > 8q2 Sol. [D] y x A(p,q) (h,0) C B O 1  cos  =   = 2 3 co-ordinates of A are       r cos  – , r sin   –  3 3            and that of B are r cos   , r sin     3 3     If p,q is the centroid of PAB 1 then p = r cos  –    r cos      2r cos  3    3 3    From equation of circle it is clealr that circle passes through origin. Let AB is chord of the circle. A  (p, q) and C(h, 0) is mid-point then B(– p + 2h,– q) and lies on the circle, x2 + y2 = px + qy we have, (– p + 2h)2 + (–q)2 = p(–p + 2h) +q(–q)  p2 + 4h2 – 4ph + q2 = – p2 + 2ph – q2  2h2 – 3ph + p2 + q2 = 0 .......(i) There are given two distinect chords which are bisceted at x-axis, then there will be two distinct values of h restisfying (i) so, discriminant of this qadratic equation must be <0 D>0  (–3p)2 – 4.2(p2 + q2) > 0  9p2 – 8p2 –8q2 > 0 p2 – 8q2 > 0  p2 > 8q2 Q.33 Sol. Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 – x + 3y = 0 on L1 and L2 are equal, then which of the following equations can represent L1 (A) x + y = 0 (B) x – y = 0 (C) x + 7y = 0 (D) x – 7y = 0 [C] Let euation of line L1 be y = mx Intercepts made by L1 and L2 on the circle will be equal i.e. L1 and L2 are at the same distacne from the centre of the circle. centre of the given circle is (1/2,–3/2), therefore 1 3 – –1 2 2 = 11  2 2 = m 3  2 2 m2  1 | m 3| 2 m2  1  8m + 8 = m2 + 6m + 9  7m2 – 6m – 1 = 0  (7m + 1) (m – 1) = 0  m = 1, –1/7 Thus two chords are x + 7y = 0 and y – x = 0 Hence x – y = 0 & x + 7y = 0 represents L1 2 Q.34 Sol. Let T1 , T2 be two tangents drawn from (–2 , 0) onto the circle C : x2 + y2 = 1. Determine the circles touching C and having T1 , T2 as their pair of tangents, Further , find the equations of all possible common tangents to these circles, when taken two at a time. y B1 T 1 A1 L 30º x M O N C1 S (–2,0) C2 A2 B2 T2 I C2 C1 From figure it is clear that triagle OLS is a right angle triangle with right angle at L. Also, OL = 1 and OS = 2 1  sin (LSO) = 2  LSO = –30º since SA1 = SA2  SA1A2 is an equilateral triangle The circle with centre at C1 is a circle Inscribed in the SA1A2. Therefore centre C1 is centroid of SA1A2 This C1 divides SM in the ratio 2 : 1 Therefore co-ordinates of C1 are (–4/3, 0) and its radius = C1M = 1/3 2 2 4  1  its equation is  x   + y2 =   .....(i) 3  3 The other circle touches the equilrteral triangle  SB1B2 extreally its radius r is given by r = s–a where B1B2 = a 1 3 But  = (a) (SN) = a 2 2 3 a and s – a = a – a = 2 2 Thus r = 3  co-ordinates of C2 are (4, 0)  equation of circle with centre at C2 is (x – 4)2 + y2 = 32 ..........(ii) Equations of common tangents to circle (i) and 1 circle are x – 1 and y = ± (x + 2) [T1 & T2] 3 Two tangents common to (i) and (ii) adre T 1 and To find the remaining two transverse tangent and (ii); we find a point I which divides the C1C2 in 1 the ratio r1 : r2 = : 3 = 1 : 9 3 51 Therefore, co-ordinates of I are (–4/5, 0) y 4  Equation of any line through I is y = m  x   5  it will touch (i) if  4 4 m –   – 0  5 5 = 1  m2 Q(3,4) (–4,3) R 1 3  39m2 = 25 = m = ± 5/ 39 P Therefore, These tangents are y=± x O 4  x   5 39  5 Q.35 The triangle PQR is inscribed in the circle x2 + y2 = 25. If Q and R have co-ordinates (3, 4) and (–4, 3) respectively, then angle QPR is equal to (A) /2 (B) /3 (C) /4 (D) /6 Sol. [C] O is the point at centre and Pis the point at circumfrecens therefore, angle QOR is double the angle QPR. so it is sufficient to find the angle QOR. Now slope of OQ = 4/3 slope of OR = –3/4 Again m1m2 = –1 Therefore, QOR = 90º which implies that QPR = 45º EXERCISE # 5 3. x2 + y2 – 4x – 6y + 9 = 0 or x2 + y2 – 20x – 22y + 121 = 0 4. (C) 5. x2 + y2 –x = 0 6. x2 + y2 + 8x –6y + 9 = 0 7. 9. k = 1 10. (A,C) 11. (A)  9 12   9  12  12.   ,  or  ,   5 5 5 5  13. x2 + y2 + gx + fy + c/2 = 0 16. x2 + y2 + 18x –2y + 32 = 0 17. 2 14. (A) 15. (C) x2 + y2 –10 x –4y + 4 = 0 18. x2 + y2 + 6x + 2y –15 = 0 and x2 + y2 –10 x –10 y + 25 = 0 19. (D) 20. a2 > 2b2  23  23.  2,   3 24. (D) 1 1 1   25. (a)  x   +  y   = (b) (–,–2)  (2,) 2 2 2     26. 7 27. (b) x2 + y2 – 1 1 28.  ,  2 4 32. (D) 33. (B, C) 22. 16x2 + 16y2 – 48x + 16y + 31= 0 21. (D) 2 2 4  34. (x – 4) 2 + y2 = 9,  x   3  30. (B) 2 + y2 = 1 (x  2) , x = ± 1, y = ± ,y=± 9 3 4  x   5 39   5 2y 3 –1 = 0 35. (C)