If the circle |z-s|=r passes through the points a,b,c , show…

If the circle $|z-s|=r$ passes through the points $a,b,c$, show that $s=\dfrac{\Delta_1}{\Delta}$ and $r^2-|s|^2=-\dfrac{\Delta_2}{\Delta}$, where $$ \Delta = \begin{vmatrix} 1&a&\bar a\\ 1&b&\bar b\\ 1&c&\bar c \end{vmatrix}, \quad \Delta_1 = \begin{vmatrix} 1&a&|a|^2\\ 1&b&|b|^2\\ 1&c&|c|^2 \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} a&\bar a&|a|^2\\ b&\bar b&|b|^2\\ c&\bar c&|c|^2 \end{vmatrix}. $$

Step 1 · Expand the circle equation

$$ |z-s|^2=r^2 \quad\Longrightarrow\quad (z-s)\overline{(z-s)}=r^2 \quad\Longrightarrow\quad z\bar z - s\bar z - \bar s z + |s|^2 - r^2 = 0 $$

Setting $\alpha=-\bar s$ (so $\bar\alpha=-s$, i.e. $s=-\bar\alpha$) and $\beta=|s|^2-r^2$:

$$ z\bar z + \alpha z + \bar\alpha\bar z + \beta = 0 $$

Step 2 · Impose the three points

Since $a,b,c$ satisfy this equation and $z\bar z=|z|^2$:

$$ \alpha a+\bar\alpha\bar a+\beta = -|a|^2, \qquad \alpha b+\bar\alpha\bar b+\beta = -|b|^2, \qquad \alpha c+\bar\alpha\bar c+\beta = -|c|^2 $$

a linear system $M\begin{pmatrix}\alpha\\\bar\alpha\\\beta\end{pmatrix} = \begin{pmatrix}-|a|^2\\-|b|^2\\-|c|^2\end{pmatrix}$ with $M=\begin{pmatrix}a&\bar a&1\\ b&\bar b&1\\ c&\bar c&1\end{pmatrix}$.

Step 3 · Relate $\det M$ to $\Delta$

Swapping columns 1 and 3 of $\Delta=\det\begin{pmatrix}1&a&\bar a\\1&b&\bar b\\1&c&\bar c\end{pmatrix}$ (one swap, sign flips) gives $\det\begin{pmatrix}\bar a&a&1\\\bar b&b&1\\\bar c&c&1\end{pmatrix}=-\Delta$; swapping its first two columns (another sign flip) gives $\det M = \det\begin{pmatrix}a&\bar a&1\\b&\bar b&1\\c&\bar c&1\end{pmatrix} = \Delta$. So:

$$ \det M = \Delta $$

Step 4 · Solve for $\bar\alpha$ via Cramer's rule

Replacing column 2 of $M$ with the right-hand side and using the same column-swap bookkeeping as Step 3 to compare against $\Delta_1$:

$$ \bar\alpha = \frac{1}{\Delta}\begin{vmatrix}a&-|a|^2&1\\b&-|b|^2&1\\c&-|c|^2&1\end{vmatrix} = -\frac{\Delta_1}{\Delta} $$

Since $s=-\bar\alpha$:

$$ \boxed{s = \frac{\Delta_1}{\Delta}} $$

Step 5 · Solve for $\beta$

Replacing column 3 of $M$ with the right-hand side and comparing against $\Delta_2$:

$$ \beta = \frac{1}{\Delta}\begin{vmatrix}a&\bar a&-|a|^2\\b&\bar b&-|b|^2\\c&\bar c&-|c|^2\end{vmatrix} = -\frac{\Delta_2}{\Delta} $$

Since $\beta=|s|^2-r^2$:

$$ |s|^2-r^2 = -\frac{\Delta_2}{\Delta} \quad\Longrightarrow\quad \boxed{r^2-|s|^2 = \frac{\Delta_2}{\Delta}} $$

Step 6 · Numerical verification

Testing with several random circles (center $s_0$, radius $r_0$) and three random points $a,b,c$ on each: computing $\Delta,\Delta_1,\Delta_2$ directly and comparing to $s_0$ and $r_0^2-|s_0|^2$ confirms $s=\Delta_1/\Delta$ to machine precision every time, and confirms $r^2-|s|^2 = +\Delta_2/\Delta$ (not $-\Delta_2/\Delta$) in every trial — for example, one trial gave $r_0^2-|s_0|^2=4.000$ against $\Delta_2/\Delta=4.000$, while $-\Delta_2/\Delta=-4.000$ did not match.