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

{\bf Question}

 Let $A$ be the Euclidean circle in ${\bf D}$ with Euclidean
center $\frac{1}{3} +\frac{1}{3}i$ and Euclidean radius
$\frac{1}{10}$.  Determine the hyperbolic center and hyperbolic
radius of $A$.
\medskip

{\bf Answer}

\begin{center}
\epsfig{file=407-8-3.eps, width=70mm}
\end{center}

The diameter of $A$ is the euclidean/hyperbolic line through 0 and
$c=\ds\frac{1}{3}(1+i)=\ds\frac{\sqrt 2}{3}e^{i\frac{\pi}{4}}$ and
so we need to consider the two points of intersection of this line
and $A$

$$p=\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10}\right)e^{i\frac{\pi}{4}}\ {\rm{and}}\
q=\left(\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10}\right)e^{i\frac{\pi}{4}}$$

\begin{eqnarray*} d_{\bf{D}}(p,q) & = &
d_{\bf{D}}\left(\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10}\right)e^{i\frac{\pi}{4}},\left(\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10}\right)e^{i\frac{\pi}{4}}\right)\\ & = &
d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10},\ds\frac{\sqrt 2}{3}-\ds\frac{1}{10}\right)\
\ \ {\rm{since}}\ m(z)=e^{\frac{-i\pi}{4}}z\ {\rm{lies\ in}}\
Isom({\bf{D}},d_{\bf{D}})\\ & = & d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10},0\right)-d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10},0\right)\end{eqnarray*}

(since $\ds\frac{\sqrt 2}{3}+\ds\frac{1}{10},\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10},0$ all lie on a hyperbolic line)

$d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10},0\right)=\ln\left(\ds\frac{1+\frac{\sqrt
2}{3}+\frac{1}{10}}{1-\frac{\sqrt
2}{3}-\frac{1}{10}}\right)=\ln\left(\ds\frac{33+10\sqrt{2}}{27-10\sqrt{2}}\right)$

$d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10},0\right)=\ln\left(\ds\frac{1+\frac{\sqrt
2}{3}-\frac{1}{10}}{1-\frac{\sqrt
2}{3}+\frac{1}{10}}\right)=\ln\left(\ds\frac{27+10\sqrt{2}}{33-10\sqrt{2}}\right)$

\newpage
So, \begin{eqnarray*} d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10},\ds\frac{\sqrt 2}{3}-\ds\frac{1}{10}\right)
& = & \ln\left(\ds\frac{33+10\sqrt{2}}{27-10\sqrt{2}}
\cdot\ds\frac{33-10\sqrt{2}}{27+10\sqrt{2}} \right)\\ & = &
\ln\left(\ds\frac{1089-200}{729-200}\right)\\ & = &
\ln\left(\ds\frac{889}{529}\right)\end{eqnarray*}

So, the hyperbolic radius of $A$ is
$\ds\frac{1}{2}\ln\left(\ds\frac{889}{529}\right)$.

The hyperbolic center is the point $ce^{\frac{i\pi}{4}}$, where
$\ds\frac{\sqrt 2}{3}-\ds\frac{1}{10}<c<\ds\frac{\sqrt
2}{3}+\ds\frac{1}{10}$ and \un{$d_{\bf{D}}\left(\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10},c\right)=\ds\frac{1}{2}\ln\left(\ds\frac{889}{529}\right)=0.25955$}

$$d_{\bf{D}}(0,c)-d_{\bf{D}}\left(0,\ds\frac{\sqrt
2}{3}-\ds\frac{1}{10}\right)=\ds\frac{1}{2}\ln\left(\ds\frac{889}{529}\right)$$

$$d_{\bf{D}}(0,c)=\ds\frac{1}{2}\ln\left(\ds\frac{889}{529}\right)+
\ln\left(\ds\frac{27+10\sqrt 2}{33-10\sqrt
2}\right)=\alpha=1.039$$

\un{$c=\tanh\left(\ds\frac{1}{2}\alpha\right)$}

\hspace{1in}
$=\ds\frac{e^{\frac{1}{2}\alpha}-e^{-\frac{1}{2}\alpha}}
{e^{\frac{1}{2}\alpha}+e^{\frac{1}{2}\alpha}} =
\ds\frac{e^{\alpha}-1}{e^{\alpha}+1}=0.4775$

\end{document}