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