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

Circle PYQ — JEE Advanced 2012 · Shift 2 · Q33 with Solution

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Correct answer: A

For the circle centered at the origin, the radius $OP$ has slope

m_{OP}=\frac{1}{\sqrt{3}}.

So the tangent at $P$ is perpendicular to $OP$ , hence its slope is

m_{PT}=-\sqrt{3}.

Since $L$ is perpendicular to $PT$ , its slope is

m_L=\frac{1}{\sqrt{3}}.

Thus any such line has form

y=\frac{x}{\sqrt{3}}+c.

Rewriting,

x-\sqrt{3}y+k=0

for some constant $k$ .

This circle has center $(3,0)$ and radius $1$ .

Let the line be

x-\sqrt{3}y+k=0.

For it to be tangent to the circle, the distance from the center $(3,0)$ to the line must equal $1$ :

\frac{|3+k|}{\sqrt{1+( -\sqrt{3})^2}}=1.

Since

\sqrt{1+3}=2,

we get

\frac{|3+k|}{2}=1 \implies |3+k|=2.

So,

3+k=\pm 2.

Hence

k=-1 \quad \text{or} \quad k=-5.

Therefore possible equations of $L$ are

x-\sqrt{3}y-1=0 \quad \text{or} \quad x-\sqrt{3}y-5=0.

That is,

x-\sqrt{3}y=1 \quad \text{or} \quad x-\sqrt{3}y=5.

Among the given options, only

x-\sqrt{3}y=1

appears.

So the correct option is A.

Stored correct answer: A

This matches our derived answer.