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

• Correct answer
  A

  $5 \sqrt{3}$
• B

  $\sqrt{65}$
• C

  10
• D

  $3 \sqrt{11}$

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 $$ \begin{array}{lll} \therefore & A B^2=B C^2-A C^2=100-25=75 \\ \Rightarrow & A B=5 \sqrt{3} \end{array} $$

Hence, the correct answer is option: 'A'.