If A(1, 1),B(-1, 1),and C(-1, -1) are three p | vedclass

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(B) Let the coordinates of point $P$ be $(x, y)$.Given points are $A(1, 1)$,$B(-1, 1)$,and $C(-1, -1)$.The condition is $PA^2 = PB^2 + PC^2$.Using the

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$x^2 + y^2 + 6x + 2y + 2 = 0$

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(B) Let the coordinates of point $P$ be $(x, y)$.Given points are $A(1, 1)$,$B(-1, 1)$,and $C(-1, -1)$.The condition is $PA^2 = PB^2 + PC^2$.Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:$PA^2 = (x - 1)^2 + (y - 1)^2$$PB^2 = (x + 1)^2 + (y - 1)^2$$PC^2 = (x + 1)^2 + (y + 1)^2$Substituting these into the condition:$(x - 1)^2 + (y - 1)^2 = [(x + 1)^2 + (y - 1)^2] + [(x + 1)^2 + (y + 1)^2]$$(x^2 - 2x + 1) + (y^2 - 2y + 1) = (x^2 + 2x + 1 + y^2 - 2y + 1) + (x^2 + 2x + 1 + y^2 + 2y + 1)$$x^2 + y^2 - 2x - 2y + 2 = 2x^2 + 2y^2 + 4x + 4$Rearranging the terms to one side:$x^2 + y^2 + 6x + 2y + 2 = 0$Thus,the equation of the locus is $x^2 + y^2 + 6x + 2y + 2 = 0$.

If $A(1, 1)$,$B(-1, 1)$,and $C(-1, -1)$ are three points and a point $P(x, y)$ moves such that $PA^2 = PB^2 + PC^2$,then the equation of the locus of $P$ is:

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