If A (c, 0) and B (-c, 0) are two points,find | vedclass

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(C) Let the coordinates of point $P$ be $(x, y)$.Given $A = (c, 0)$ and $B = (-c, 0)$.$PA^{2} = (x - c)^{2} + (y - 0)^{2} = x^{2} - 2cx + c^{2} + y^{2

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

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(C) Let the coordinates of point $P$ be $(x, y)$.Given $A = (c, 0)$ and $B = (-c, 0)$.$PA^{2} = (x - c)^{2} + (y - 0)^{2} = x^{2} - 2cx + c^{2} + y^{2}$.$PB^{2} = (x + c)^{2} + (y - 0)^{2} = x^{2} + 2cx + c^{2} + y^{2}$.$AB^{2} = (c - (-c))^{2} + (0 - 0)^{2} = (2c)^{2} = 4c^{2}$.Given the condition $PA^{2} + PB^{2} = AB^{2}$,we substitute the expressions:$(x^{2} - 2cx + c^{2} + y^{2}) + (x^{2} + 2cx + c^{2} + y^{2}) = 4c^{2}$.$2x^{2} + 2y^{2} + 2c^{2} = 4c^{2}$.$2x^{2} + 2y^{2} = 2c^{2}$.Dividing by $2$,we get $x^{2} + y^{2} = c^{2}$.

If $A (c, 0)$ and $B (-c, 0)$ are two points,find the locus of point $P$ such that $PA^{2} + PB^{2} = AB^{2}$.

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