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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 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 x−√3y=1The equation of tangent PT to circle x2+y2=4 at the point P(√3,1) is √3x+y=4. Let, slope of line L be m. m×(−√3)=−1(∵ L is perpendicular to PT) m=1√3 Any line with slope 1√3 will be of the form y=1√3x+c Given that this line is a tangent to the circle (x−3)2+y2=1 ⇒∣1√3−0+c√(1√3)2+(−1)2∣=1 ⇒c+√32=±1√3⇒c=−√3±2√3 Equation of line 'L' is y=1√3x=√3±2√3 √3y=x−3±2 ⇒3y=x−5 or √3y=x−1

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