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(D) Let the circumcenter be $O(x, y)$. The distance from $O$ to each vertex is equal,so $OA^2 = OB^2 = OC^2$.$OA^2 = (x-6)^2 + (y-0)^2 = x^2 - 12x + 3

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(D) Let the circumcenter be $O(x, y)$. The distance from $O$ to each vertex is equal,so $OA^2 = OB^2 = OC^2$.$OA^2 = (x-6)^2 + (y-0)^2 = x^2 - 12x + 36 + y^2$$OB^2 = (x-0)^2 + (y-6)^2 = x^2 + y^2 - 12y + 36$$OC^2 = (x-7)^2 + (y-7)^2 = x^2 - 14x + 49 + y^2 - 14y + 49 = x^2 + y^2 - 14x - 14y + 98$Equating $OA^2 = OB^2$:$x^2 - 12x + 36 + y^2 = x^2 + y^2 - 12y + 36$$-12x = -12y \implies x = y$Equating $OB^2 = OC^2$:$x^2 + y^2 - 12y + 36 = x^2 + y^2 - 14x - 14y + 98$Since $x = y$,substitute $y$ with $x$:$x^2 + x^2 - 12x + 36 = x^2 + x^2 - 14x - 14x + 98$$2x^2 - 12x + 36 = 2x^2 - 28x + 98$$16x = 62 \implies x = \frac{62}{16} = 3.875$Since $x = y$,the circumcenter is $(3.875, 3.875)$.None of the given options match this result.

If the coordinates of the vertices of triangle $ABC$ are $(6, 0)$,$(0, 6)$,and $(7, 7)$ respectively,find the circumcenter of the triangle.

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