The locus of the point $P$ such that $(P A)^{2} + (P B)^{2} = 10$ where $A (2, 0)$ and $B (4, 0)$ is:

x^{2} + y^{2} + 6 x + 5 = 0

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x^{2} + y^{2} - 6 x + 5 = 0

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x^{2} + y^{2} - 6 x - 5 = 0

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x^{2} + y^{2} + 6 x - 5 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 6 x + 5 = 0$
$(h - 2)^{2} + (k - 0)^{2} + (h - 4)^{2} + (k - 0)^{2} = 10$
$h^{2} + k^{2} - 6 h + 5 = 0$
$x^{2} + y^{2} - 6 x + 5 = 0$

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