The circumcenter (center of circumscribed circle) of the triangle $ ABC $ is point:
$$ \left(\dfrac{ 15 }{ 2 },~\dfrac{ 1205 }{ 72 }\right) $$explanation
The circumcenter of a triangle is given by:
$$ C = \left( ~ \frac{\begin{vmatrix} A_x^2 + A_y^2 & A_y & 1 \\ B_x^2 + B_y^2 & B_y & 1 \\ C_x^2 + C_y^2 & C_y & 1 \\ \end{vmatrix}} {2 \cdot \begin{vmatrix} A_x & B_y & 1 \\ B_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~,~ \frac{\begin{vmatrix} A_x & A_x^2 + A_y^2 & 1 \\ B_x & B_x^2+B_y^2 & 1 \\ C_x & C_x^2+C_y^2 & 1 \\ \end{vmatrix}} {2 \cdot \begin{vmatrix} A_x & B_y & 1 \\ A_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~ \right) $$where $ A_x $ and $ A_y $ are $ x $ and $ y $ coordinates of the point $ A $ , $ B_x $ and $ B_y $ are $ x $ and $ y $ coordinates of the point $ B $ etc..
In this example we have : $ A_x = -\frac{ 5 }{ 3 } $ , $ A_y = \frac{ 245 }{ 9 } $ , $ B_x = 15 $ , $ B_y = 5 $ , $ C_x = 0 $ and $ C_y = 5 $ .
When we insert these values into the formula, we obtain the given result.