Q. If the circle $x^{2} + y^{2} + 6x - 2y + k = 0$ bisects the circumference of the circle $x^{2} + y^{2} + 2x - 6y - 15 = 0$, then $k$ is equal to
A) 21
B) $- 21$
C) 23
D) $- 23$
Explanation
(d)
The condition for a circle bisecting the circumference of the second circle is
$2g_{2}\left( g_{1} - g_{2} \right) + 2f_{2}\left( f_{1} - f_{2} \right) = c_{1} - c_{2}$
$\Rightarrow 2(1)(3 - 1) + 2( - 3)( - 1 + 3) = k + 15$
$\Rightarrow 2(2) + ( - 6)(2) = k + 15$
$\Rightarrow 4 - 12 = k + 15$
$\Rightarrow \ - 8 = k + 15$
$\Rightarrow k = - 23$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.