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\begin{document}

{\bf Question}

In the proof that the group of M\"obius transformations acts
transitively on the set of circles in $\overline{\bf C}$, we use
the fact that three distinct points in $\overline{\bf C}$
determine a circle in $\overline{\bf C}$.

\medskip
\noindent Prove this fact in a single special case: determine the
center and radius of the circle determined by $2 +i$, $3 -i$, and
$-7i$.
\medskip

{\bf Answer}

The circle in $\bf {C}$ determined by $2+i, 3-i, -7i$ (first note
that these 3 points do not lie on a line in $\bf {C}$, sine the
line through $2+i$ and $3-i$ intersects the imaginary axis at $5i
\ne -7i$).

\begin{itemize}
\item
The midpoint of the line segment through $2+i$ and $3-i$ is
$\ds\frac{1}{2}(2+i+3-i)=\ds\frac{5}{2}$ and its slope is
$\ds\frac{-1-1}{3-2}=-2$.

The perpendicular bisector then has equation $y-0=
+\ds\frac{1}{2}\left(x-\ds\frac{5}{2}\right)$

$\Rightarrow \un{y=\ds\frac{1}{2}x-\ds\frac{5}{4}}$

\item
The midpoint of the line segment through $2+i$ and $-7i$ is
$\ds\frac{1}{2}(2+i-7i)=1-3i$ and its slope is
$\ds\frac{-7-1}{0-2}=4$.

The perpendicular bisector then has equation
$y+3=-\ds\frac{1}{4}(x-1)=-\ds\frac{1}{4}x+\ds\frac{1}{4}$

$\Rightarrow \un{y=-\ds\frac{1}{4}x-\ds\frac{11}{4}}$
\end{itemize}

These perpendiculars are both diameters of the desired circle and
hence intersect at the center:

\begin{eqnarray*} \ds\frac{1}{2}x-\ds\frac{5}{4} & = &
-\ds\frac{1}{4}x-\ds\frac{11}{4}\\ \ds\frac{3}{4}x & = &
-\ds\frac{6}{4} \end{eqnarray*}

\un{$x=-2$}

\un{$y=\ds\frac{1}{2}x-\ds\frac{5}{4}=-\ds\frac{9}{4}$}

So, the center of the circle is \un{$-2-\ds\frac{9}{4}i=a$}.

\un{Its radius is}:

$\begin{array} {rcl} |\ -2-\frac{9}{4}i-(-7i)| & = &
|-2+\frac{19}{4}i|=5.154\\ |-2-\frac{9}{4}i-(2+i)| & = &
|-4-\frac{13}{4}i|=5.154\\ |-2-\frac{9}{4}i-(3-i)| & = &
|-5-\frac{5}{4}i\ |=5.154 \end{array}$
\end{document}
