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.)