Where is the Center?

Let the vertices be $A,B,C,D$ where $AB = 1, BC = 4, CD = 8, DA = 7$ . Reflect $D$ over the angle bisector of $AC$ to $D'$ . Observe that $D'$ is still on the circle. We also have $CD' = 7$ and $AD' = 8$ . Consider quadrilateral $ABCD'$ . We know that $AD'^2 + AB^2 = CD'^2 + BC^2 = 65$ which implies that they must be right triangles. This further implies that the diameter must be the square root of the sum which is $\sqrt{65}$ .