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+2√2y=6
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Solution
The correct option is Dx+2√2y=6 Here, 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+m2or3m=−√1+m2⇒m2=1+m2or9m2=1+m2⇒mϵϕorm=±12√2∴y=±12√2x±2√1+18⇒y=±x2√2±62√2⇒2√2y=±(x+6)∴x+2√2y=6