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Sum of the squares of the distances from a point to (c,0) and (-c,0) is $4 c^{2}$ . It's locus is

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$x^{2} + y^{2} = 4 c^{2}$

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

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

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

answer is C.

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Detailed Solution

$A (c, 0) B (- c, 0) .$ Let the point be $P (x, y)$

$P A^{2} + P B^{2} = 4 c^{2} \Rightarrow {(x - c)}^{2} + y^{2} + {(x + c)}^{2} + y^{2} = 4 c^{2} \Rightarrow 2 x^{2} + 2 y^{2} = 2 c^{2} \Rightarrow x^{2} + y^{2} = c^{2}$

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