If the circumference of the circle `x^2 + y^2 + 8x + 8y - b = 0` is bisected by the circle `x^2 + y^2 - 2x + 4y + a = 0` then `a+b=` (A) 50 (B) 56 (C) `-56` (D) `-34`

Options

A. 50
B. 56
C. `-56`
D. `-34`

Answer

3

Explanation

The equation of radical axis ( i.e., common chord ) of the two circles is
`10x+4ya-a-b=0` (1)
The center of the first circle is `H(-4,-4)`
Since the second circle bisects the circumference of the first circle, the center `H(-54,-4)` of the first circle must lie on the coomon chord (1). Therefore, ltbr. `-40-16-1-b=0`
or `a+b= -56`