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)
Equation of the common chord is $S_{1} - S_{2} = 0$
$\therefore\ \left( x^{2} + y^{2} + 6x - 2y + k \right) - \left( x^{2} + y^{2} + 2x - 6y - 15 \right) = 0$
$\Rightarrow 4x + 4y + k + 15 = 0$
Centre of second circle is $C_{2}( - 1,\ 3)$
Since, equation of the chord passes through the centre $( - 1,\ 3)$ of second circle
$\therefore\ \ 4( - 1) + 4(3) + k + 15 = 0\ \ \Rightarrow \ \ k = - 23$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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