Mathematics - Circle Question with Solution | TestHub

Equation of tangent $PT$ 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 $PT$ be $x-y\sqrt{3}+\lambda =0$ As it is tangent to the circle $(x-3{)}^2+y^2=1$ $\therefore$ 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 $x-\sqrt{3}y=5$