The equation of tangent to the circle \(x^2+y^2=64\) at the point \(\mathrm{P}\left(\frac{2 \pi}{3}\right)\) is

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
\(
(\mathrm{y}-4 \sqrt{3})=\frac{1}{\sqrt{3}}(\mathrm{x}+4) \Rightarrow \mathrm{x}-\sqrt{3} \mathrm{y}+16=0
\)