A point P(x, y) is such that the sum of squar | vedclass

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(C) The distance of point $P(x, y)$ from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$.The sum of squares of these distances is $x^2 + y^2$.The

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$x^2+y^2+2xy+2x-2y-1=0$

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(C) The distance of point $P(x, y)$ from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$.The sum of squares of these distances is $x^2 + y^2$.The distance of point $P(x, y)$ from the line $x-y-1=0$ is given by $d = \frac{|x-y-1|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y-1|}{\sqrt{2}}$.According to the problem,the sum of squares of distances from the axes is equal to the square of the distance from the line:$x^2 + y^2 = \left( \frac{|x-y-1|}{\sqrt{2}} \right)^2$$x^2 + y^2 = \frac{(x-y-1)^2}{2}$$2(x^2 + y^2) = x^2 + y^2 + 1 - 2xy - 2x + 2y$$2x^2 + 2y^2 = x^2 + y^2 + 1 - 2xy - 2x + 2y$$x^2 + y^2 + 2xy + 2x - 2y - 1 = 0$.

$A$ point $P(x, y)$ is such that the sum of squares of its distances from the coordinate axes is equal to the square of its distance from the line $x-y=1$. Then the equation of the locus of $P$ is

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