ABCD is a square with side length a. Taking A | vedclass

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(B) Let the vertices of the square be $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.Since the circle passes through these four points,it must satisfy th

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$x^2 + y^2 - ax - ay = 0$

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(B) Let the vertices of the square be $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.Since the circle passes through these four points,it must satisfy the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $A(0, 0)$,we get $c = 0$.Substituting $B(a, 0)$,we get $a^2 + 2ga = 0 \Rightarrow g = -a/2$.Substituting $D(0, a)$,we get $a^2 + 2fa = 0 \Rightarrow f = -a/2$.Substituting these values into the general equation,we get $x^2 + y^2 - ax - ay = 0$.Alternatively,the circle has diameter $AC$ with endpoints $(0, 0)$ and $(a, a)$. The equation is $(x-0)(x-a) + (y-0)(y-a) = 0$,which simplifies to $x^2 - ax + y^2 - ay = 0$ or $x^2 + y^2 - ax - ay = 0$.

$ABCD$ is a square with side length $a$. Taking $AB$ and $AD$ as the coordinate axes,find the equation of the circle passing through the vertices of the square.

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