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

Here, tangent to $x^2+y^2=4$ at $(\sqrt{3},1)$ is $\sqrt{3}x+y=4$ .... (i)

As L is perpendicular to $\sqrt{3}x+y=4$
$\Rightarrow x-\sqrt{3}y=\lambda$ which is tangent to
$(x-3{)}^2+y^2=1$
$\Rightarrow \frac{|3-0-\lambda |}{\sqrt{1+3}}=1\left(Distanceofcentreofcirclefromtangent\right)\Rightarrow |3-\lambda |=2\Rightarrow 3-\lambda =2,-2\therefore \lambda =1,5\Rightarrow L:x-\sqrt{3}y=1,x-\sqrt{3}y=5$