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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 : 50 of 60

Marks: +1, -0

Solution:

Point of intersection of direct common tangents is (6, 0).

Let the equation of common tangent be

y - 0 = m(x - 6)

as it touches

x^2 + y^2

= 4.

⇒

\left|\frac{0-0+6m}{\sqrt{1+m^2}}\right|

= 2

9m^2

1+m^2

m =

\pm \frac{1}{2\sqrt{2}}

So, the equation of common tangent is

y =

\frac{1}{2\sqrt{2}}

(x - 6), y =

-\frac{1}{2\sqrt{2}}(x-6)

and also x = 2.

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