If a circle $S$ with radius 5 touches the circle $x^2+y^2-6 x-4 y-12=0$ at $(-1,-1)$, then the length of the…
If a circle $S$ with radius 5 touches the circle $x^2+y^2-6 x-4 y-12=0$ at $(-1,-1)$, then the length of the tangent from the centre of the circle $S$ to the given circle is
$5 \sqrt{3}$
$\sqrt{65}$
10
$3 \sqrt{11}$
Solution
Given circle is
$
\begin{aligned}
& x^2+y^2-6 x-4 y-12=0 \\
& \Rightarrow \quad(x-3)^2+(y-2)^2=25 \\
&
\end{aligned}
$
According to the question,
$\because \quad \triangle C A B$ is right angle triangle, so
$
\begin{array}{lll}
\therefore & A B^2=B C^2-A C^2=100-25=75 \\
\Rightarrow & A B=5 \sqrt{3}
\end{array}
$
