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

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Circle

A tangent

P T

is drawn to the circle

x^{2} + y^{2} = 4

at the point

P (3, 1)

. A straight line

L

, perpendicular to

P T

is a tangent to the circle

(x - 3)^{2} + y^{2} = 1

.

A possible equation of

L

(A) $x - 3 y = 1$ ✓ Correct
(B) $x + 3 y = 1$
(C) $x - 3 y = - 1$
(D) $x + 3 y = 5$

Answer: (A)

Solution

The tangent at

P (3, 1)

x^{2} + y^{2} = 4

is:

3 x + y = 4

Its slope is

- 3

, so a line perpendicular to it has slope

\frac{1}{3}

.

Tangents to

(x - 3)^{2} + y^{2} = 1

with slope

\frac{1}{3}

are:

y = \frac{1}{3} (x - 3) \pm 1 + \frac{1}{3} = \frac{1}{3} (x - 3) \pm \frac{2}{3}

Simplifying:

3 y = x - 3 \pm 2

So:

x - 3 y = 1

x - 3 y = 5

.

Option (A) matches.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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