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 $(x-3{)}^2+y^2=1$. $\left(1\right)$ 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)

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

Slope of tangent to $x^2+y^2=4$ at $P(\sqrt{3},1)$ is given by differentiating given equation w.r.t. x

$\frac{dy}{dx}=\frac{-x}{y}=-\sqrt{3}$

So slope of line L $=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$

Let equation of L be $x-\sqrt{3}y+c=0$

As L is tangent to $(x-3{)}^2+y^2=1$

So perpendicular distance of L from its center should be equal to its radius i.e. $1$.

Centre $(3,0)$

$\frac{|3+c|}{\sqrt{4}}=1$

$\Rightarrow c=-1or-5$

So equation of L is $x-\sqrt{3}y=1$

or

$x-\sqrt{3}y=5$

Hence option A is correct.