Let a circle passing through (2, 0) have its centre at the point (h, k). Let (xc, yc ) be the point of intersection of the lines 3x + 5y = 1 and (2 + c)x + 5c2y = 1. If h = c 1 xc and k = c 1 yc, then the equation of the circle is :
- 5x2 + 5y2 - 4x - 2y - 12 = 0
- 25x2 + 25y2 - 20x + 2y - 60 = 0
- 25x2 + 25y2 - 2x + 2y - 60 = 0
- 5x2 + 5y2 - 4x + 2y - 12 = 0
Answer
(2) 25x2 + 25y2 - 20x + 2y - 60 = 0
Solution
Multiply 3x + 5y = 1 by c2 and subtract from the second line:
(2 + c - 3c2 )x = 1 - c2 xc = (1 - c)(1 + c)(1 - c)(2 + 3c) = 1 + c2 + 3c
yc = 1 - 3xc5 = -15(2 + 3c)
So h = 25 and k = - 125
Since the circle passes through (2, 0): r2 = (2 - 25 )2 + ( 125 )2 = 1601625
Circle: (x - 25 )2 + (y + 125 )2 = 1601625
25x2 + 25y2 - 20x + 2y - 60 = 0
Answer: (2)