The extremities of a diagonal of a rectangle are $(0, 0)$ and $(4, 4)$ . The locus of the extremities of the other diagonal is equal to

x^{2} + y^{2} - 4 x - 4 y = 0

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x^{2} + y^{2} + 4 x + 4 y - 4 = 0

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x^{2} + y^{2} - 4 x - 4 y + 4 = 0

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x^{2} + y^{2} - 4 x - 4 y - 4 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 4 x - 4 y = 0$
REF.Image.
MP of AC = MP of BD
MP of BD $= (\frac{4}{2}, \frac{4}{2})$
= (2,2)
AC = BD
$↓$
OC = OD
$O D = \sqrt{(4 - 2)^{2} + (4 - 2)^{2}} = 2 \sqrt{2}$
$(x - 2)^{2} + (y - 2)^{2} = 8$
$x^{2} - 4 x + 4 + y^{2} - 4 y + y = 8.$
$x^{2} + y^{2} - 4 x - 4 y = 0$

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