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

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(B) $ABCD$ is a square with side $16$ units. Let $a = 16$ units.The vertices are $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.The centre of the circle

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$4$

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(B) $ABCD$ is a square with side $16$ units. Let $a = 16$ units.The vertices are $A(0, 0)$,$B(a, 0)$,$C(a, a)$,and $D(0, a)$.The centre of the circle circumscribing the square is the midpoint of the diagonal $AC$,which is $O\left(\frac{a}{2}, \frac{a}{2}\right)$.The diameter of the circle is the diagonal $AC = \sqrt{(a-0)^2 + (a-0)^2} = a\sqrt{2}$.Thus,the radius $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.The equation of the circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.Substituting $h = \frac{a}{2}$,$k = \frac{a}{2}$,and $r = \frac{a}{\sqrt{2}}$:$\left(x-\frac{a}{2}\right)^2 + \left(y-\frac{a}{2}\right)^2 = \left(\frac{a}{\sqrt{2}}\right)^2$$x^2 - ax + \frac{a^2}{4} + y^2 - ay + \frac{a^2}{4} = \frac{a^2}{2}$$x^2 + y^2 - ax - ay = 0$$x^2 + y^2 = a(x+y)$Given $a = 16$,the equation is $x^2 + y^2 = 16(x+y)$.Comparing this with $x^2 + y^2 = 4k(x+y)$,we get $4k = 16$,so $k = 4$.

$ABCD$ 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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