ABCD is a square with side 16 units and A is | vedclass

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(B) The vertices of the square are $A(0,0)$,$B(16,0)$,$C(16,16)$,and $D(0,16)$.Since the circle circumscribes the square,the diagonal $AC$ is the diam

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$4$

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(B) The vertices of the square are $A(0,0)$,$B(16,0)$,$C(16,16)$,and $D(0,16)$.Since the circle circumscribes the square,the diagonal $AC$ is the diameter of the circle.The coordinates of $A$ are $(0,0)$ and $C$ are $(16,16)$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the coordinates of $A$ and $C$:$(x-0)(x-16) + (y-0)(y-16) = 0$$x(x-16) + y(y-16) = 0$$x^2 - 16x + y^2 - 16y = 0$$x^2 + y^2 = 16(x+y)$Given the equation $x^2 + y^2 = 4k(x+y)$,we compare the two equations:$4k = 16$$k = 4$

$ABCD$ is a square with side $16$ units and $A$ is the origin. If the equation of the circle circumscribing the square $ABCD$ is $x^2+y^2=4k(x+y)$,then $k=$

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