Prop. If \(0,1,z,w\) all lie on a circle centered at \(1/2\) then \(zw\) is the orthogonal projection of \(0\) onto the line through \(z\) and \(w\).
Proof.
If \(z,w\) to lie on that circle, then
\[(z-\frac12)(\overline{z}-\frac12)=\frac14,\text{ which simplifies to }z\overline{z}=\frac{z+\overline{z}}2\]
\[(w-\frac12)(\overline{w}-\frac12)=\frac14,\text{ which simplifies to }w\overline{w}=\frac{w+\overline{w}}2.\]
Introducing names \(z=a+bi,w=c+di\) these equations turn into
\[a^2+b^2=a,\quad c^2+d^2=c.\]
Two things need to be shown: \((1)\) \(zw\) lies on the line through \(z\) and \(w\), and \((2)\) \(zw\) is perpendicular to \(z-w\).
\((1)\) Verify that the following determinant is zero.
\begin{align*}
\begin{vmatrix}1&a&b\\1&c&d\\1&ac-bd&ad+bc\end{vmatrix}&=acd+bc^2-acd+bd^2-a^2d-abc+abc-b^2d+ad-bc\\
&=b(c^2+d^2)-(a^2+b^2)d+ad-bc=bc-ad+ad-bc=0.
\end{align*} \((2)\) Verify that \(zw/(z-w)\) is imaginary, or equivalently that \(zw(\overline z-\overline w)\) is imaginary.
\begin{align*}
zw(\overline z-\overline w)+\overline{zw(\overline z-\overline w)}&=z\overline zw-zw\overline w+z\overline z\overline w-\overline zw\overline w=z\overline z(w+\overline w)-(z+\overline z)w\overline w\\
&=\frac{z+\overline z}2(w+\overline w)-(z+\overline z)\frac{w+\overline w}2=\frac12(z+\overline z)(w+\overline w)(1-1)=0.
\end{align*}
