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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 ' $k$ ' is equal to
Options:
Solution:
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Verified Answer
The correct answer is:
–23
Given, equation of circles are
$$
\begin{aligned}
& S: x^2+y^2+6 x-2 y+k=0 \\
& S^{\prime}: x^2+y^2+2 x-6 y-15=0
\end{aligned}
$$
Centre $=(-1,3)$
Equation of common chaod is $S-S^{\prime}=0$
$$
4 x+4 y+(k+15)=0
$$
Centre of $S^{\prime}$ lies on the above line $\therefore(-1,3)$ lies on Eq. (i)
$$
4(-1)+4(3)+k+15=0 \Rightarrow k=-23
$$
Hence, option (4) is correct.
$$
\begin{aligned}
& S: x^2+y^2+6 x-2 y+k=0 \\
& S^{\prime}: x^2+y^2+2 x-6 y-15=0
\end{aligned}
$$
Centre $=(-1,3)$
Equation of common chaod is $S-S^{\prime}=0$
$$
4 x+4 y+(k+15)=0
$$
Centre of $S^{\prime}$ lies on the above line $\therefore(-1,3)$ lies on Eq. (i)
$$
4(-1)+4(3)+k+15=0 \Rightarrow k=-23
$$
Hence, option (4) is correct.
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