Point P is on the orthogonal hyperbola $x^{2} - y^{2} = a^{2}$ . Point P' is the perpendicular projection of P on the x-axis. Then, $| P P^{'} |^{2}$ is equal to the power of point P' relative to which circle?

x^{2} + y^{2} = a^{2}

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x^{2} + y^{2} = a^{2} + b^{2}

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Director circle

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Auxiliary circle

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Solution

The correct options are
A $x^{2} + y^{2} = a^{2}$
B Auxiliary circle
$P (a s e c (t), a t a n (t))$ and $P^{'} (a s e c (t), 0)$
$| P P^{'} | = a^{2} t a n^{2} (t)$
The power of point P' relative to a circle $x^{2} + y^{2} = a^{2}$ is :
$(a s e c (t) - 0)^{2} + (0 - 0)^{2} - a^{2}$ (power of the point w.r.t. circle)
$= a^{2} s e c^{2} (t) - a^{2} = a^{2} (s e c^{2} (t) - 1) = a^{2} t a n^{2} (t)$
Hence, the correct options are A and D.

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