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 $x-\sqrt{3}y=1$

The equation of tangent PT to circle $x^2+y^2=4$ at the point $P(\sqrt{3},1)$ is $\sqrt{3}x+y=4$.
Let, slope of line L be m.
$m\times (-\sqrt{3})=-1(\because$ L is perpendicular to PT)
$m=\frac{1}{\sqrt{3}}$
Any line with slope $\frac{1}{\sqrt{3}}$ will be of the form
$y=\frac{1}{\sqrt{3}}x+c$
Given that this line is a tangent to the circle $(x-3{)}^2+y^2=1$
$\Rightarrow \mid \frac{\frac{1}{\sqrt{3}}-0+c}{\sqrt{(\frac{1}{\sqrt{3}}{)}^2+(-1{)}^2}}\mid =1$
$\Rightarrow \frac{c+\sqrt{3}}{2}=\pm \frac{1}{\sqrt{3}}\Rightarrow c=-\sqrt{3}\pm \frac{2}{\sqrt{3}}$
Equation of line 'L' is
$y=\frac{1}{\sqrt{3}}x=\sqrt{3}\pm \frac{2}{\sqrt{3}}$
$\sqrt{3}y=x-3\pm 2$
$\Rightarrow 3y=x-5$ or $\sqrt{3}y=x-1$