\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

For each point $p$ in ${\bf H}$, $p\ne i$, determine the equation
of the Euclidean circle or line containing the hyperbolic line
through $p$ and $i$, in terms of ${\rm Re}(p)$ and ${\rm Im}(p)$.
\medskip

{\bf Answer}

\un{If Re($p$)=0}, then the hyperbolic line through $p$ and $i$
has the equation $\{\rm{Re}(z)=0\}$. (So a vertical euclidean
line.)

\un{If Re($p$)=0}, the slope of the euclidean line segment through
$p$ and $i$ is $m=\ds\frac{\rm{Im}(p)-1}{\rm{Re}(p)}$ and the
midpoint is $\ds\frac{1}{2}(p+i)$. So, the perpendicular bisector
has the equation
$$y-\ds\frac{1}{2}(\rm{Im}(p)+1)=\ds\frac{\rm{Re}(p)}{1-\rm{Im}(p)}
\left(x-\ds\frac{1}{2}\rm{Re}(p)\right).$$

Setting $y=0$ and solving for $x$ we see that the euclidean circle
containing the hyperbolic line through $i$ and $p$ has \un{center}
$a$

\begin{eqnarray*} a & = &
-\ds\frac{1}{2}(\rm{Im}(p)+1)\ds\frac{(1-\rm{Im}(p))}{\rm{Re}(p)}+
\ds\frac{1}{2}\rm{Re}(p)\\ & = &
\ds\frac{-1+\rm{Im}(p)^2}{2\rm{Re}(p)}+\ds\frac{\rm{Re}(p)^2}
{2\rm{Re}(p)}=\ds\frac{|p|^2-1}{2\rm{Re}(p)}.
\end{eqnarray*}
The \un{radius} of the circle is:

$$r=\left|\ds\frac{\rm{Im}(p)^2}{2\rm{Re}(p)}-i\right|=
\sqrt{\left(\ds\frac{|p|^2-1}{2\rm{Re}(p)}\right)^2+1}$$

\end{document}