Q46 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
The length of the tangent from $(6,8)$ to the circle $x^2+y^2=4$ is
- A$\sqrt{6}$
- B$2 \sqrt{6}$
- Correct answerC$4 \sqrt{6}$
- D$5 \sqrt{6}$
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.
Hence, the correct answer is option: 'C'.