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If the radius of a circle $x^{2}+y^{2}-4 x+6 y-k=0$ is 5, then $\mathrm{k}=$
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2680 Upvotes
Verified Answer
The correct answer is:
$12$
Given equation of circle is
$$
\begin{array}{l}
x^{2}+y^{2}-4 x+6 y-k=0 \\
r \quad=\sqrt{4+9+k} \Rightarrow 5=\sqrt{13+k} \Rightarrow 13+k=25 \\
k=12
\end{array}
$$
$$
\begin{array}{l}
x^{2}+y^{2}-4 x+6 y-k=0 \\
r \quad=\sqrt{4+9+k} \Rightarrow 5=\sqrt{13+k} \Rightarrow 13+k=25 \\
k=12
\end{array}
$$
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