The equation of tangent to the circle $x^2+y^2=64$ at the point $\mathrm{P}\left(\frac{2 \pi}{3}\right)$ is
The equation of tangent to the circle $x^2+y^2=64$ at the point $\mathrm{P}\left(\frac{2 \pi}{3}\right)$ is
$x-\sqrt{3} y-16=0$
$\sqrt{3} \mathrm{x}+\mathrm{y}-16=0$
$\mathrm{x}+\sqrt{3} \mathrm{y}+16=0$
$\mathrm{x}-\sqrt{3} \mathrm{y}+16=0$
Solution
Circle $x^2+y^2=(8)^2$, has radius 8 and centre $(0,0)$. Point $\mathrm{P}\left(\frac{2 \pi}{3}\right)$ on the circle has coordinates
$\mathrm{P} \equiv\left(8 \cos \frac{2 \pi}{3}, 8 \sin \frac{2 \pi}{3}\right) \text { i.e. } \mathrm{P} \equiv(-4,4 \sqrt{3})$
Differentiating equation of circle w.r.t. $\mathrm{x}$, we get
$2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{y} \Rightarrow\left(\frac{d y}{d x}\right)_P=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}$
Hence required equation of tangent is
$(\mathrm{y}-4 \sqrt{3})=\frac{1}{\sqrt{3}}(\mathrm{x}+4) \Rightarrow \mathrm{x}-\sqrt{3} \mathrm{y}+16=0$