Finding center and radius of circumcircle

Find the center and radius of the circle which circumscribes the triangle with (complex) vertices $a_1,a_2,a_3$. Express the result in symmetric form.

I'm not sure where to start in this question. The circumcenter is the intersection of the perpendicular bisectors, but I don't know how to calculate the line equation of the perpendicular bisectors yet. As for the radius, after we find the circumcenter $c$ we can calculate it using $|c-a_1|$ (or $|c-a_2|$ or $|c-a_3|$; all three should be equal.) So how can I calculate the circumcenter?

Here is a way of getting started. It might not be the best way, but it makes significant progress. Let $c$ be the centre of the circle and $r$ (a real number) the radius. You can write an equation for the square of the radius as $$(a_1-c)(\bar{a}_1-\bar c)=r^2=|a_1|^2+|c|^2-c\bar a_1-a_1\bar c$$

[note this is just the cosine rule for triangles in different notation]

You can write two other equations like this, and use them to eliminate $r^2$, $|c|^2$ (which appear together) and $\bar c$ so that you have a linear equation for $c$ in terms of things you know.

Let $R$ be the radius of the circumcircle of $a_1,a_2,a_3$ then the circumcenter $z$ is defined by the condition $|z-a_1|=|z-a_2|=|z-a_3|=R\,$. Expanding $R^2=|z-a_1|^2=(z-a_1)(\bar z - \overline{a_1})$ and similar for $a_2,a_3$ gives the system:

$$ \begin{cases} \begin{align} \overline{a_1}\,z + a_1\,\bar z + (R^2 - |z|^2) & = |a_1|^2 \\ \overline{a_2}\,z + a_2\,\bar z + (R^2 - |z|^2) & = |a_2|^2 \\ \overline{a_3}\,z + a_3\,\bar z + (R^2 - |z|^2) & = |a_3|^2 \end{align} \end{cases} $$

Eliminating $\bar z$ and $(R^2 - |z|^2)$ between the equations amounts to solving it as a linear system in $z$,$\bar z$ and $(R^2-|z|^2)$ which, by Cramer's rule, gives:

$$ z\;=\;\frac{\left| \begin{array}{ccc} |a_1|^2 & a_1 & -1 \\ |a_2|^2 & a_2 & -1 \\ |a_3|^2 & a_3 & -1 \end{array} \right|} {\left| \begin{array}{ccc} \overline{a_1} & a_1 & -1 \\ \overline{a_2} & a_2 & -1 \\ \overline{a_3} & a_3 & -1 \end{array} \right|} \;=\;\frac{\left| \begin{array}{ccc} a_1 & |a_1|^2 & 1 \\ a_2 & |a_2|^2 & 1 \\ a_3 & |a_3|^2 & 1 \end{array} \right|} {\left| \begin{array}{ccc} a_1 & \overline{a_1} & 1 \\ a_2 & \overline{a_2} & 1 \\ a_3 & \overline{a_3} & 1 \end{array} \right|} $$

(The formula itself has been posted in another answer here. The above is a derivation thereof.)