Circles as Hermitian matrices

A traditional circle has a center $c$ and a radius $r$. We think of $c$ as a complex number, and $r$ is a positive real number. The set of all points $z$ on this circle are precisely the solutions to $$\vert z-c\vert =r$$ where we are using the absolute value of complex numbers to represent distance. If we square this equation and make use of complex conjugation, we can write it as
$$\begin{align*} 0 &= \vert z-c\vert^2 - r^2\\ &=(z-c)(\overline{z}-\overline{c}) -r^2\\ &= z \bar{z} -\bar{c} z-c \bar{z} +\vert c\vert^2-r^2 \end{align*}$$
This can be written in a convenient matrix format as $$0 = \begin{bmatrix}\bar{z} & 1\end{bmatrix} \begin{bmatrix}1... ...ert c\vert^2-r^2 \end{bmatrix}\begin{bmatrix}z\\ 1\end{bmatrix}$$ The $2\times 2$ matrix in the middle is an example of a Hermitian matrix $$H = \begin{bmatrix}a & b\\ \bar{b} & c\end{bmatrix}$$ which satisfies $H^T = \bar{H}$. Here $H^T$ is the transpose of $H$, and $\bar{H}$ is the matrix obtained by conjugating all the entries of $H$. Since $H^T = \bar{H}$, we know that $\det H^T=\det H = \bar{\det H}$, and thus that the determinant is a real number. In this case, the determinant is $\vert c\vert^2-r^2-c\bar{c}=-r^2$ is negative.