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In a square $A B C D$ of side length $a$, suppose $A B$ and $A D$ are along the coordinate axes. Then, the circle that circumscribes the square is
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Verified Answer
The correct answer is:
$x^2+y^2-a(x+y)=0$
$B D$ is diameter.
$\therefore$ Circle is
$$
\begin{array}{rr}
& (x-0)(x-a)+(y-a)(y-0)=0 \\
\Rightarrow & x^2+y^2-a(x+y)=0
\end{array}
$$
$\therefore$ Circle is
$$
\begin{array}{rr}
& (x-0)(x-a)+(y-a)(y-0)=0 \\
\Rightarrow & x^2+y^2-a(x+y)=0
\end{array}
$$
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