Cramer's rule and centre of a circle The exact problem of finding the centre of a circle circumscribing the triangle formed by three complex numbers $a_1, a_2$ and $a_3$ is considered in detail here: Finding center and radius of circumcircle. What I am interested in is knowing why the last column of the first equality in @dvix's answer if full of -1s, when the variable to be solved is $R^2 - |z|^2$. Link to the answer.
 A: To be more pedagogical, he should have had the last row be all $1$'s so as to match the variable he states i.e. $R^2 - |z|^2$ as you say, but the end result is the same - by determinant rules, you can pull out the $-1$ in both the numerator and the denominator which will result in:
$$\frac{\det\begin{bmatrix}|a_1|^2 & a_1 & -1 \\ |a_2|^2 & a_2 & -1 \\ |a_3|^2 & a_3 & -1\end{bmatrix}}{\det\begin{bmatrix}\overline{a_1} & a_1 & -1 \\ \overline{a_2} & a_2 & -1 \\ \overline{a_3} & a_3 & -1\end{bmatrix}} = \frac{-\det\begin{bmatrix}|a_1|^2 & a_1 & 1 \\ |a_2|^2 & a_2 & 1 \\ |a_3|^2 & a_3 & 1\end{bmatrix}}{-\det\begin{bmatrix}\overline{a_1} & a_1 & 1 \\ \overline{a_2} & a_2 & 1 \\ \overline{a_3} & a_3 & 1\end{bmatrix}} = \frac{\det\begin{bmatrix}|a_1|^2 & a_1 & 1 \\ |a_2|^2 & a_2 & 1 \\ |a_3|^2 & a_3 & 1\end{bmatrix}}{\det\begin{bmatrix}\overline{a_1} & a_1 & 1 \\ \overline{a_2} & a_2 & 1 \\ \overline{a_3} & a_3 & 1\end{bmatrix}}$$
The reason is, that it doesn't matter whether his variable is $R^2 - |z|^2$ or $|z|^2 - R^2$, the latter will just change the system slightly by having minuses in front of the variable, since $R^2 - |z|^2 = -(|z|^2 - R^2)$. That is:
$$\begin{cases}
\begin{align}
\overline{a_1}\,z + a_1\,\bar z + (R^2 - |z|^2) & = |a_1|^2  \\
\overline{a_2}\,z + a_2\,\bar z + (R^2 - |z|^2) & = |a_2|^2  \\
\overline{a_3}\,z + a_3\,\bar z + (R^2 - |z|^2) & = |a_3|^2
\end{align}
\end{cases}$$
Is the same as:
$$\begin{cases}
\begin{align}
\overline{a_1}\,z + a_1\,\bar z - (|z|^2 - R^2) & = |a_1|^2  \\
\overline{a_2}\,z + a_2\,\bar z - (|z|^2 - R^2) & = |a_2|^2  \\
\overline{a_3}\,z + a_3\,\bar z - (|z|^2 - R^2) & = |a_3|^2
\end{align}
\end{cases}$$
So it seems like he used $|z|^2 - R^2$ as a variable to setup his solution. It's clear however, that this doesn't change the solution for $z$.
