$A (1, 2), B (2, - 3), C (- 2, 3)$ are three points. A point P moves such that $P A^{2} + P B^{2} = 2 P C^{2}$ Find the locus of P.

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Solution

Let point $P$ be $(x, y)$

Given $A (1, 2), B (2, - 3), C (- 2, 3)$ and

$P A^{2} + P B^{2} = 2 P C^{2}$

$⟹ (\sqrt{(x - 1)^{2} + (y - 2)^{2}})^{2} + (\sqrt{(x - 2)^{2} + (y + 3)^{2}})^{2} = 2 (\sqrt{(x + 2)^{2} + (y - 3)^{2}})^{2}$

$⟹ x^{2} + 1 - 2 x + y^{2} + 4 - 4 y + x^{2} + 4 - 4 x + y^{2} + 9 + 6 y = 2 (x^{2} + 4 + 4 x + y^{2} + 9 - 6 y)$

$⟹ 2 x^{2} + 2 y^{2} - 6 x + 2 y + 18 = 2 x^{2} + 2 y^{2} + 8 x - 12 y + 26$

$⟹ 14 x - 14 y + 8 = 0$

$⟹ 7 x - 7 y + 4 = 0$

Therefore, the locus of point $P$ is $7 x - 7 y + 4 = 0$

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