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(C) The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$. The vertices of the rectangle are $A(-2, -2)$,$B(6, -2)$,$C(6, 5)$,and $D(-2, 5)

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$x^2+y^2-4x-3y-22=0$

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(C) The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$. The vertices of the rectangle are $A(-2, -2)$,$B(6, -2)$,$C(6, 5)$,and $D(-2, 5)$. The diagonal of the rectangle acts as the diameter of the circle. Taking diagonal $AC$ with endpoints $A(-2, -2)$ and $C(6, 5)$,the equation of the circle in diameter form is given by: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ $(x - (-2))(x - 6) + (y - (-2))(y - 5) = 0$ $(x + 2)(x - 6) + (y + 2)(y - 5) = 0$ $x^2 - 6x + 2x - 12 + y^2 - 5y + 2y - 10 = 0$ $x^2 + y^2 - 4x - 3y - 22 = 0$

If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$,then the equation of the circle,drawn on the diagonal of this rectangle as its diameter,is

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