If the circle$x^{2} + y^{2} + 4x + 22y + c = 0\ \$bisects the circumference of the circle$x^{2} + y^{2} - 2x + 8y - d = 0$, then $c + d$ is equal to

Question

Dr. Abubacker · Accepted Answer

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

Q. If the circle$x^{2} + y^{2} + 4x + 22y + c = 0\ \$bisects the circumference of the circle$x^{2} + y^{2} - 2x + 8y - d = 0$, then $c + d$ is equal to

Explanation

Reviewed by Dr. Abubacker

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