If the length of the tangent drawn from the point from $(-2,3)$ to the circle $x^2+y^2+8 x$ $-6 y+k=0$ is 4…
If the length of the tangent drawn from the point from $(-2,3)$ to the circle $x^2+y^2+8 x$ $-6 y+k=0$ is 4 units, then $k$ is equal to
34
36
38
37
Solution
Circle $C: x^2+y^2+8 x-6 y+k=0$
Length of tangents drawn by an external points to the circle is $\sqrt{S_1}$.
Now, the length of tangent drawn by $(-2,3)$ to the circle is $L$,
$
\begin{aligned}
& & L & =\sqrt{S_1} \\
& & L & =\sqrt{(-2)^2+(3)^2+8(-2)-6(3)+\mathrm{k}} \\
\Rightarrow & & 4 & =\sqrt{4+9-16-18+k} \text { where } L=4 \\
\Rightarrow & & 4 & =\sqrt{k-21} \Rightarrow 16=k-21 \\
\Rightarrow & & k & =16+21 \Rightarrow k=37
\end{aligned}
$