Given $A=(a_1,a_2), B=(b_1,b_2), C=(c_1,c_2)$, is there a simple formula to express the radius of the circumcircle of $ABC$?
Note that you could compute the radius from the sidelengths as $\frac{abc}{\sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}}$, but I'm really hoping that there's something simpler than that. If it helps, assume $C=(0,0)$.
The square root in the denominater is equal to the twice the area of $\triangle ABC$ (Heron's formula). The same area can be calculated by embeding the plane in 3-dimensional space and using the vector product: $$ S = \frac12 |\vec{CA} \times \vec{CB}| $$ Using Cartesian coordinates, we have $$ |\vec{CA} \times \vec{CB}| = |(a_1-c_1)(b_2-c_2)-(a_2-c_2)(b_1-c_1)| = \\ = |a_1(b_2-c_2)+b_1(c_2-a_2)+c_1(a_2-b_2)| $$ so $$ R = \frac{\sqrt{(a_1-c_1)^2+(a_2-c_2)^2}\sqrt{(b_1-c_1)^2+(b_2-c_2)^2}\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}}{|a_1(b_2-c_2)+b_1(c_2-a_2)+c_1(a_2-b_2)|}$$