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Question

The locus of the point of intersection of two perpendicular tangents to the circle x2+y2=a2 is

A
x2+y2=a22
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B
x2+y2=a23
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C
x2+y2=2a2
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D
x2+y2=3a2
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Solution

The correct option is C x2+y2=2a2
Equation of a tangent to the circle x2+y2=a2 is y = mx + a1+m2 (1)
Equation of a tangent to the circle x2+y2=a2 and perpendicular to (1) is
y = (1m) x + a 1+(1m) y = - xm + a1 + m2m my = - x + a1+m2 (2)
From (1), y - mx = a1+m2
From (2), my + x = a1+m2
Squaring and adding the above questions, we get (ymx)2+(my+x)2=a2(1+m2)+a2(1+m2)
(1+m2)x2+(1+m2)y2=2a2(1+m2)
x2+y2=2a2 which is the required locus.

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