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If the circle $x^2+y^2+6 x-2 y+k=0$ bisects the circumference of the circle $x^2+y^2+2 x-6 y-15=0$, then $k=$
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Verified Answer
The correct answer is:
-23
Equation of given circles
$$
\begin{array}{r}
S_1: x^2+y^2+6 x-2 y+k=0 \\
\text { and } S_2: x^2+y^2+2 x-6 y-15=0
\end{array}
$$
Since, the circle $S_1$ bisects the circumference of the circle $S_2$, then the common of circles $S_1$ and $S_2$ will passing through the centre of circle $S_2(-1,3)$.
$\because$ Equation of common chord of circles $S_1$ and $S_2$ is
$\because$ Chord (i) is passes through point $(-1,3)$, so $-4+12+k+15=0 \Rightarrow k=-23$
Hence, option (c) is correct.
$$
\begin{array}{r}
S_1: x^2+y^2+6 x-2 y+k=0 \\
\text { and } S_2: x^2+y^2+2 x-6 y-15=0
\end{array}
$$
Since, the circle $S_1$ bisects the circumference of the circle $S_2$, then the common of circles $S_1$ and $S_2$ will passing through the centre of circle $S_2(-1,3)$.
$\because$ Equation of common chord of circles $S_1$ and $S_2$ is
$\because$ Chord (i) is passes through point $(-1,3)$, so $-4+12+k+15=0 \Rightarrow k=-23$
Hence, option (c) is correct.
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