AP EAMCET · Maths · Circle
The length of the tangent from \((6,8)\) to the circle \(x^2+y^2=4\) is
- A \(\sqrt{6}\)
- B \(2 \sqrt{6}\)
- C \(4 \sqrt{6}\)
- D \(5 \sqrt{6}\)
Answer & Solution
Correct Answer
(C) \(4 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Let \(P=(6,8)\) \[ S: x^2+y^2-4=0 \] \[ \begin{aligned} \text { Length of tangent } & =\sqrt{S_{11}}=\sqrt{(6)^2+(8)^2-4} \\ & =\sqrt{36+64-4}=\sqrt{96}=4 \sqrt{6} \end{aligned} \] Hence, option (3) is correct.
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