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(A) Let the point be $P(x, y)$.The sum of the squares of the distances from $(0,0), (1,0), (0,1), (1,1)$ is given by:$(x^2 + y^2) + ((x-1)^2 + y^2) +

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$16$

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(A) Let the point be $P(x, y)$.The sum of the squares of the distances from $(0,0), (1,0), (0,1), (1,1)$ is given by:$(x^2 + y^2) + ((x-1)^2 + y^2) + (x^2 + (y-1)^2) + ((x-1)^2 + (y-1)^2) = 18$Expanding the terms:$(x^2 + y^2) + (x^2 - 2x + 1 + y^2) + (x^2 + y^2 - 2y + 1) + (x^2 - 2x + 1 + y^2 - 2y + 1) = 18$Combining like terms:$4x^2 + 4y^2 - 4x - 4y + 4 = 18$Dividing by $4$:$x^2 + y^2 - x - y + 1 = 4.5$$x^2 + y^2 - x - y - 3.5 = 0$The radius $r$ of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$.Here,$2g = -1 \Rightarrow g = -0.5$,$2f = -1 \Rightarrow f = -0.5$,and $c = -3.5$.$r = \sqrt{(-0.5)^2 + (-0.5)^2 - (-3.5)} = \sqrt{0.25 + 0.25 + 3.5} = \sqrt{4} = 2$.The diameter $d = 2r = 2(2) = 4$.Therefore,$d^2 = 4^2 = 16$.

The locus of a point,which moves such that the sum of squares of its distances from the points $(0,0), (1,0), (0,1), (1,1)$ is $18$ units,is a circle of diameter $d$. Then $d^{2}$ is equal to ...... .

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