Question

Let a circle C touch the lines L1 : 4x – 3y +K1 = 0 and L2 : 4x – 3y + K2 = 0, K1, K2∈R. If a line passing through the centre of the circle C intersects L1 at (–1, 2) and L2 at (3, –6), then the equation of the circle C is :

• A. (x – 1)2 + (y – 2)2 = 4
• B. (x + 1)2 + (y – 2)2 = 4
• C. (x – 1)2 + (y + 2)2 = 16
• D. (x – 1)2 + (y – 2)2 = 16

Answer 1

Answer: (C)

(x – 1)2 + (y + 2)2 = 16

Answer 2

Co-ordinate of centre,
$C=(\frac {3+(−1)}{2},\frac {−6+2}{2})$
$C = (1,−2)$
L 1 is passing through A,
–4 – 6 + K 1 = 0
K 1 = 10
L 2 is passing through B,
12 + 18 + K 2 = 0
K 2 = –30
Equation of L 1: 4 x – 3 y + 10 = 0
Equation of L 1: 4 x – 3 y – 30 = 0
Diameter of circle
=$|\frac {10+30}{\sqrt {4^2+(−3)^2}}|=8$
So, The radius = 4
Equation of circle (x – 1)2 + (y + 2)2 = 16

Hence, the correct option is (C): (x – 1)2 + (y + 2)2 = 16