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If $A (b, 0), B (- b, 0)$ then the locus of the point P such that $P A^{2} + P B^{2} = 2 c^{2}$ is

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$x^{2} + y^{2} + b^{2} + c^{2} = 0$

$2 x^{2} + y^{2} + 3 b^{2} - c^{2} = 0$

$x^{2} - y^{2} - b^{2} - c^{2} = 0$

$x^{2} + y^{2} + b^{2} - c^{2} = 0$

answer is A.

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Detailed Solution

$\begin{array}{l} A (b, 0), B (- b, 0), P (x, y) \\ P A^{2} + P B^{2} = 2 c^{2} \\ (x - b)^{2} + y^{2} + (x + b)^{2} + y^{2} = 2 c^{2} \\ x^{2} + y^{2} + b^{2} - c^{2} = 0 \end{array}$

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