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IIT JEE Advanced 2012 Paper 2

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A tangent PT is drawn to the circle

x^2 + y^2

= 4 at the point P( 3, 1). A straight line L, perpendicularto PT is a tangent to the circle

(x - 3)^2 + y^2

= 1.

Section: Mathematics

Question : 49 of 60

Marks: +1, -0

Solution:

The equation of tangent at

P(\sqrt{3},1)

\sqrt{3} x + y = 4

Slope of line perpendicular to above tangent is

\frac{1}{\sqrt{3}}

So, the equation of tangents with slope

\frac{1}{\sqrt{3}}

(x-3)^2 + y^2

= 1 will be,

y =

\frac{1}{\sqrt{3}} (x-3) \pm 1 \sqrt{1+\frac{1}{3}}

\sqrt{3} y

= x - 3 ± (2)

\sqrt{3} y

= x - 1 or

\sqrt{3} y

= x - 5

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