In a square ABCD of side length a,suppose AB | vedclass

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(B) The vertices of the square are $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.Since the square is inscribed in a circle,the diagonal $AC$ (or $BD$) i

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$x^2+y^2-a(x+y)=0$

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(B) The vertices of the square are $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.Since the square is inscribed in a circle,the diagonal $AC$ (or $BD$) is the diameter of the circle.Using the diameter form of the circle equation with endpoints $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (a, a)$:$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$(x - 0)(x - a) + (y - 0)(y - a) = 0$$x(x - a) + y(y - a) = 0$$x^2 - ax + y^2 - ay = 0$$x^2 + y^2 - a(x + y) = 0$

In a square $ABCD$ of side length $a$,suppose $AB$ and $AD$ are along the coordinate axes. Then,the circle that circumscribes the square is

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