IIT-JEE 2012 · Paper 2 · Mathematics Q53

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MathematicsSingle correct+3 / -1Question 53 of 60

Passage

Paragraph for Questions 53 and 54

A tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt3,1)$ . 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

x-\sqrt3y=1

x+\sqrt3y=1

x-\sqrt3y=-1

x+\sqrt3y=5

Answer and worked solution

Official answer: A

Tangent at P. $\sqrt3x+y=4$ , slope $-\sqrt3$ .

L. Perpendicular, slope $\frac1{\sqrt3}$ : $x-\sqrt3y=c$ . Tangent to the circle of centre $(3,0)$ , radius 1:
$\frac{|3-c|}2=1\implies c=1\text{ or }5.$
So $L$ is $x-\sqrt3y=1$ or $x-\sqrt3y=5$ ; the first is listed.

(A)

Topics:Tangents and normals Distance of a point from a line

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IIT-JEE 2012 · Paper 2, Mathematics question 53