AP EAMCET · Maths · Circle
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
- A \(5 \sqrt{3}\)
- B \(\sqrt{65}\)
- C 10
- D \(3 \sqrt{11}\)
Answer & Solution
Correct Answer
(A) \(5 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
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…
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