Q .     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

JEE 2012 ADVANCED Paper - 2
A

`x -\sqrt{3}y=1`

B

`x +\sqrt{3}y=1`

C

`x-\sqrt{3}y=-1`

D

`x+\sqrt{3}y=5`

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