22. Fint the equation of the circle whose centre is on y‐axis and which passes through the points (‐2,1) and (1,3). 23. Find the length of the common chord of the circles x2+y2 ‐2x‐4 = 0 and x2+y2+2x+4y‐6 = 0 SOLUTIONS 1 . ÷ by 2 Centre = (‐g , ‐f) = 2. 3x2+3y2‐6x+ky+1=0 ÷ by 3 s.B.s and simplifying we get k = ± 4 3. let the other end be ( x , y) , given centre = (1,‐3) mid point of (x,y) and (‐4 , 5) = (1 , ‐3) X = 6 and y = ‐11. The other end is (6 , ‐11). 4. parametric equations are X = ‐3 +
24 cos
and y = ‐4 + 24 sin
Because centre = (‐3 , ‐4) and r = 5. centre = (‐1, 2) and r = 24 6 Equation of the circle is (x − (− 1)) + ( y − 2)
2
2
= 6 i.e. x2+y2+2x‐4y‐1=0 6. equation of the circle is [x + 2][x + 3] + [ y − 3][ y − 5] = 0 i.e. x2+y2+5x‐8y+21 = 0 7. m=tan600= 3 .;. tangent will be of the form y = 3 x+c The condition for aline to be a tangent is c2 = a2(1+m2) Substituting for a2 = 9 and m = 3 we get c = ±6 The equations of tangents are y = 3x±6 8. centre (1 , ‐2) , it touches y‐axis .:. f2 = c = 4. Equation of the circle is x2 +y2 ‐2x +4y+4 = 0 9. PT = (− 1) + (− 3)
2
2
− 2(− 1) − (− 3) − 7 = 8 = 2 2
units 10. power of the point (‐1 , ‐2) w.r.t to the given circle = 1+4‐3‐6+1 = ‐3 < 0 . power of the point (0 , 1) = 0+1+0+3+1 = 5 > 0 .:. (‐1 , ‐2) lies inside and (0 , 1) lies outside the circle. 11. x2+y2 +6x+2y‐1=0 …………(1) and 2x2+2y2+6y‐3 = 0 ÷2 x2+y2+3y‐1 = 0 ……..(2) Equation of the common tangent of (1) and (2) Is the radical axis i.e. (1) –(2) i.e. 6x‐y+1/2 = 0 12x – 2y +1 = 0 TWO MARKS; 12 . CENTRE = (3, ‐1) = (‐g , ‐f) .;. the equation of the circle will be x2+y2+6x‐2y+c = 0 Since it passes through (7 , 4) 49+16‐42‐8+c = 0 Î c = ‐31 .;. the equation of the circle is x2+y2+6x‐2y‐31 = 0 13. centre = intersection of the diameters x+2y = 4 , x+y = 6 = (8 , ‐2) Given r = 10 .;. the equation of the circle is x2+y2 ‐16x+4y‐32 = 0 14 . tangent is parallel to the line 5x +12y – 3 =0. .:. it will be of the form 5x + 12y +k = 0. Length of the perpendicular from the centre (1 , ‐1) to the line = radius Î k −7
= 3 Î k‐7 = ± 39 13
.:. k = 46 or ‐32 .:. The equations of tangents are 5x+2y+46 = 0 and 5x+2y‐32 = 0 . 15 . let P(x1 , y1) be any point on the locus. .:. power of the po int w.r.t. x 2 + y 2 = 6
2
= 2
2
power of the po int w.r.t. x + y + 3x + 3 y = 0 1
X12+y12‐6 = 2( X12+y12+3x1+3y1) Simplifying we get x2+y2+4x+4y+2 = 0 as the equation of the locus. 16. Here g = 2, f = ‐3, c = 4 Intercept on x‐axis = 2 g 2 − c = 0 Intercept on y‐axis = 2 f 2 − c =2 17. R.A. of (1) & (2) is 2x – 2 = 0 i.e. x ‐ 1 = 0 R.A. of (2) & (3) is 4x – y +(5/2) = 0 Solving we get x = 1, y = 13/2 .:. Radical centre = (1, 13/2) 18. condition 2g1 g2 +2f1f2 = c1+c2 .’. 2(‐1)(k/2) + 2(3/2)(1/2) = ‐3/2 + 0 Î k = 3 19. Perpendicular distance from} (2, ‐3) to the line } = Radius .’. 3.2 − 4. − 3 − 8
9 + 16
= r 2 = r .’. Equation of the circle is (x‐2)2 + (y+3)2 = 2 2 2
2
i.e. x +y ‐4x+6y+9 = 0 20. Centre of circle (1) = C1 = (‐2,0) Centre of circle (2) = C2 = (6, 6) Radius of circle (1) = r1 = 7 Radius of circle (2) = r2 = 3 C1 C2 = 10 .’. C1 C2 = r1 + r2 Circles touch each other externally Point of contact divides C1 C2 internally in the ratio 7:3 .’. P = ( 18/5, 21/5) 21. g = 2, f= ‐3, c= ‐3 Centre = (‐2, 3) r = 4 p = length of the perpendicular from (‐2,3) to the line x+ y = 5 p = 2 √2 length of the required chord = 2
r 2 − p2
4√ 2 units 22. Here g = 0 2
2 .’. let the circle be x +y + 2fy + c = 0 Since it passes through (‐2,1) and (1,3) We have 2f + c = ‐5 and 6f +c = ‐10 Solving, these we get f = ‐5/4 and c = ‐5/2 .’. equation of the circle is 2x2+2y2‐5y ‐5 = 0 . 23. R.A of the given circles is 4x+4y ‐ 2 = 0 i.e. 2x+2y ‐ 1 = 0 centre of First circle = (1, 0 ) the radius of First circle = r =√5 p = perpendicular distance from (1, 0) to 2x+2y‐1 = 0 = 1
8
Length of the common chord = 2
r 2 − p2
=
39
2
units ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐