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AP EAMCET · Maths · Circle

The length of the tangent from \((6,8)\) to the circle \(x^2+y^2=4\) is

  1. A \(\sqrt{6}\)
  2. B \(2 \sqrt{6}\)
  3. C \(4 \sqrt{6}\)
  4. D \(5 \sqrt{6}\)
Verified Solution

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.