Circumscribed Circle of Triangle
Input: $(x,y)=$ coordinates, triangle vertices numbered $(0,1,2)$.
Output: $(a,b)=$ midpoint, $R=$ radius.
$$
\begin{cases}
(x_0-a)^2+(y_0-b)^2=R^2 \\
(x_1-a)^2+(y_1-b)^2=R^2 \\
(x_2-a)^2+(y_2-b)^2=R^2
\end{cases}
$$
$$
\begin{cases}
x_0^2-2x_0a+a^2 + y_0^2-2y_0b+b^2 = R^2 \\
x_1^2-2x_1a+a^2 + y_1^2-2y_1b+b^2 = R^2 \\
x_2^2-2x_2a+a^2 + y_2^2-2y_2b+b^2 = R^2
\end{cases}
$$
$$
\begin{cases}
(x_1^2-x_0^2)-2(x_1-x_0)a + (y_1^2-y_0^2)-2(y_1-y_0)b = 0 \\
(x_2^2-x_0^2)-2(x_2-x_0)a + (y_2^2-y_0^2)-2(y_2-y_0)b = 0
\end{cases}
$$
$$
\begin{bmatrix}
x_1-x_0 & y_1-y_0 \\ x_2-x_0 & y_2-y_0
\end{bmatrix}
\begin{bmatrix} a \\ b \end{bmatrix} = \frac{1}{2}
\begin{bmatrix}
(x_1^2-x_0^2) + (y_1^2-y_0^2) \\ (x_2^2-x_0^2) + (y_2^2-y_0^2)
\end{bmatrix}
$$
$$
R = \sqrt{(x_0-a)^2+(y_0-b)^2}
$$