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Question

A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT is a tangent to the circle (x3)2+y2=1
A possible equation of L is


A
x3y=1
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B
x+3y=1
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C
x3y=1
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D
x+3y=5
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Solution

The correct option is A x3y=1

Here, tangent to x2+y2=4 at (3,1) is 3x+y=4 .... (i)


As L is perpendicular to 3x+y=4
x3y=λ which is tangent to
(x3)2+y2=1
|30λ|1+3=1 (Distance of centre of circle from tangent)|3λ|=23λ=2,2λ=1,5L:x3y=1,x3y=5


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