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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 (x−3)2+y2=1 A common tangent of the two circles is

A
x=4
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B
y=2
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C
x+3y=4
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D
x+22y=6
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Solution

## The correct option is D x+2√2y=6Here, equation of common tangent be y=mx±2√1+m2 Which is also the tangent to (x−3)2+y2=1 ⇒ |3m−0+2√1+m2|√m2+1=1⇒ 3m+2√1+m2=±√1+m2⇒ 3m=−3√1+m2or 3m=−√1+m2⇒ m2=1+m2 or 9m2=1+m2⇒mϵϕ or m=±12√2∴ y=±12√2x±2√1+18⇒ y=±x2√2±62√2⇒ 2√2y=±(x+6)∴ x+2√2y=6

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