A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$,…
A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^{2}+y^{2}=1$.
Question: A possible equation of $L$ is
$x-\sqrt{3} y=1$
$x+\sqrt{3} y=1$
$x-\sqrt{3} y=-1$
$x+\sqrt{3} y=5$
Solution
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 $x-\sqrt{3} y=5$