If the sum of the distances from a variable p | vedclass

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(C) Let the point $P$ be $(x, y)$.Given that $AP + BP = 2$.$\sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y-1)^2} = 2$.$\sqrt{(x-1)^2 + y^2} = 2 - \sqrt{x^2 + (

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$3x^2+2xy+3y^2-4x-4y=0$

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(C) Let the point $P$ be $(x, y)$.Given that $AP + BP = 2$.$\sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y-1)^2} = 2$.$\sqrt{(x-1)^2 + y^2} = 2 - \sqrt{x^2 + (y-1)^2}$.Squaring both sides:$(x-1)^2 + y^2 = 4 + x^2 + (y-1)^2 - 4\sqrt{x^2 + (y-1)^2}$.$x^2 - 2x + 1 + y^2 = 4 + x^2 + y^2 - 2y + 1 - 4\sqrt{x^2 + (y-1)^2}$.$-2x + 2y - 4 = -4\sqrt{x^2 + (y-1)^2}$.Dividing by $-2$:$x - y + 2 = 2\sqrt{x^2 + (y-1)^2}$.Squaring both sides again:$(x - y + 2)^2 = 4(x^2 + y^2 - 2y + 1)$.$x^2 + y^2 + 4 - 2xy + 4x - 4y = 4x^2 + 4y^2 - 8y + 4$.$3x^2 + 2xy + 3y^2 - 4x - 4y = 0$.

If the sum of the distances from a variable point $P$ to the given points $A(1,0)$ and $B(0,1)$ is $2$,then the locus of $P$ is

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