A tangent P T is drawn to the circle x^ 2 +y^ 2 =4 at the point P(...

Equation of tangent $P T$ to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$ is $x \sqrt{3}+y=4$ Let the line $L$, perpendicular to tangent $P T$ be $x-y \sqrt{3}+\lambda=0$ As it is tangent to the circle $(x-3)^{2}+y^{2}=1 \therefore \quad$ Length of perpendicular from centre of circle to the Tangent = radius of circle. $\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text{ or }-5 \therefore$ Equation of $L$ can be $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$