Question

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 $(x-3)^{2}+y^{2}=1$
A possible equation of $L$ is

• A. $x-\sqrt{3} y=1$
• B. $x+\sqrt{3} y=1$
• C. $x-\sqrt{3} y=-1$
• D. $x+\sqrt{3} y=5$

Answer 1

Answer: (A)

Equation of tangent at $P(\sqrt{3}, 1)$
$\sqrt{3} x+y=4$
Slope of line perpendicular to above tangent is $\frac{1}{\sqrt{3}}$
So equation of tangents with slope $\frac{1}{\sqrt{3}}$ to $(x-3)^{2}+y^{2}=1$ will be
$y=\frac{1}{\sqrt{3}}(x-3) \pm 1 \sqrt{1+\frac{1}{3}}$
$\sqrt{3} y=x-3 \pm(2)$
$\sqrt{3} y=x-1$ or $\sqrt{3} y=x-5$