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Question

Point P is on the orthogonal hyperbola x2y2=a2. Point P' is the perpendicular projection of P on the x-axis. Then, |PP|2 is equal to the power of point P' relative to which circle?

A
x2+y2=a2
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B
x2+y2=a2+b
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C
Director circle
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D
Auxiliary circle
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Solution

The correct options are
B x2+y2=a2
D Auxiliary circle

P(asec(t),atan(t)) and P(asec(t),0)
|PP|=a2tan2(t)
The power of point P' relative to a circle x2+y2=a2 is :

(asec(t)0)2+(00)2a2 (power of the point w.r.t. circle)

=a2sec2(t)a2=a2(sec2(t)1)=a2tan2(t)

Hence, the correct options are A and D.

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