Explanation

(b)
Given that, circle $S_{1} \equiv x^{2} + y^{2} + 4x + 22y + c = 0$ bisects the circumference of the circle
$S_{2} = x^{2} + y^{2} - 2x + 8y - d = 0$
The common chord of the given circles is
$S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} + 4x + 22y + c - x^{2} - y^{2} + 2x - 8y + d = 0$
$\Rightarrow 6x + 14y + c + d = 0$ …(i)
So, Eq. (i) passes through the centre of the second circle, $ie,\ (1,\ - 4)$
$\therefore\ 6 - 56 + c + d = 0$
$\Rightarrow c + d = 50$