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Let $S$ be the circle in $x y$-plane which touches the $x$-axis at point $A$, the $y$-axis at point $B$ and the unit circle $x^{2}+y^{2}=1$ at point $C$ externally. If $O$ denotes the origin, then the angle OCA equals
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$\frac{5 \pi}{8}$
$\therefore$ Slope of $O M=1$ $\therefore \angle \mathrm{COA}=45^{\circ}$ $\mathrm{C}\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Now OM $=O C+C M$
$\sqrt{2} \mathrm{~h}=1+\mathrm{h}$
Squaring
$\begin{array}{l}
2 h^{2}=h^{2}+1+2 h \\
h^{2}-2 h-1=0
\end{array}$
$\mathrm{h}=\frac{2 \pm \sqrt{8}}{2} \Rightarrow \mathrm{h}=1 \pm \sqrt{2}$
$\mathrm{~A}(1+\sqrt{2}, 0)$
Slope of $A C=\frac{0-\frac{1}{\sqrt{2}}}{1+\sqrt{2}-\frac{1}{\sqrt{2}}}=-\frac{1}{\sqrt{2}+1}=-(\sqrt{2}-1)$
$=-\tan 22 \frac{1}{2}^{\circ}=\tan \left(180-22 \frac{1}{2}\right)^{\circ}=\tan 157 \frac{1}{2}^{\circ}$
$\therefore \angle \mathrm{CAX}=157 \frac{1}{2}^{\circ}$
$\therefore \angle \mathrm{OCA}=157 \frac{1}{2}^{\circ}-45^{\circ}=112 \frac{1}{2}^{\circ}=\frac{5 \pi}{8}$
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