Question

A tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3},1).$ A straight line $L,$ perpendicular to $PT$ 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

The correct option is A $x-\sqrt{3}y=1$


Slope of $PT=\tan \left({120}^{\circ }\right)=-\sqrt{3}$
Slope of line $L\text{is}\frac{1}{\sqrt{3}}$
Line $L\equiv x-\sqrt{3}y+\lambda =0$
$L$ isTangent to $(x-3{)}^2+y^2=1$
$\Rightarrow \frac{|3+\lambda |}{2}=1$
$\Rightarrow \lambda +3=\pm 2,$
$\Rightarrow \lambda =-1,-5$
$\therefore L\equiv x-\sqrt{3}y-1=0$ or $x-\sqrt{3}y-5=0$