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

The length of the tangent from $(6,8)$ to the circle $x^2+y^2=4$ is
  1. $\sqrt{6}$
  2. $2 \sqrt{6}$
  3. $4 \sqrt{6}$
  4. $5 \sqrt{6}$

Solution

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

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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