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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Determine the hyperbolic line in the upper half plane that passes
through the points $3 + i$ and $-2 + 4i$.

\medskip

{\bf Answer}

Set $p=3+i$ and $q=-2+4i$. Since Re($p$)$\ne$Re($q$),the
hyperbolic line through $p$ and $q$ lies in a euclidean circle of
some center $a$ and some radius $r$.
\begin{itemize}
\item
calculate details of the (euclidean) line segment joining $p-q$:

slope is
$m=\ds\frac{\rm{Im}(p)-\rm{Im}(q)}{\rm{Re}(p)-\rm{Re}(q)}=-\ds\frac{3}{5}$

midpoint is $\ds\frac{1}{2}(p+q)=\ds\frac{1}{2}+\ds\frac{5}{2}i$

\item
calculate equation of perpendicular bisector:

slope is $-\ds\frac{1}{m}=\ds\frac{5}{3}$ through
$\ds\frac{1}{2}+\ds\frac{5}{2}i$ and so its equation is

$y-\ds\frac{5}{2}=\ds\frac{5}{3}\left(x-\ds\frac{1}{2}\right)$

\item
intersection with $x$-axis occurs at $(a,0)$:

$-\ds\frac{5}{2}=\ds\frac{5}{3}\left(a-\ds\frac{1}{2}\right)$. So
$-\ds\frac{3}{2}=a-\ds\frac{1}{2}$ and \underline{$a=-1$}.

\item
radius is $|a-p|=|q-a|$:

$|a-p|=|-1-3-i|=|-4-i|=\sqrt{17}$

$|a-q|=|-1+2-4i|=|1-4i|=\sqrt{17}$

(as a check).

So, this hyperbolic line is contained in the euclidean circle with
center $a=-1$ and radius $r=\sqrt{17}$.
\end{itemize}

\end{document}