23. A circle is inscribed in a semicircle with centre $O$ and diameter $A B$. The centre of the circle is the point $P$, where $P A=P O$.
What is the ratio of the radius of the circle to the radius of the semicircle?

A $4: 9$
В $3: 8$
C $3: 7$
D $2: 5$
E $1: 2$
23. B Let the radii of the semicircle and circle be $R$ and $r$ respectively; let the perpendicular from $P$ to $A O$ meet $A O$ at $C$. So $C$ is the point where $A B$ touches the circle. Also let the semicircle and the circle touch at the point $D$.
The radii $O D$ and $P D$ are both perpendicular to the common tangent at $D$, so $O P D$ is a straight line. Hence $O P$ has length $R-r$.
Consider triangles $A C P$ and $O C P: P A=P O$ (given); $\angle A C P=\angle O C P=90^{\circ} ; P C$ is common to both triangles. So the triangles are congruent (RHS) and we deduce that $A C=O C$.
Hence $O C=\frac{O A}{2}=\frac{R}{2}$.
Applying Pythagoras' Theorem to triangle $O C P$ :
$O P^{2}=O C^{2}+C P^{2}$. Therefore $(R-r)^{2}=\left(\frac{R}{2}\right)^{2}+r^{2}$.
So $R^{2}-2 R r+r^{2}=\frac{R^{2}}{4}+r^{2}$.
Hence $\frac{3 R^{2}}{4}=2 R r$, that is $r=\frac{3 R}{8}$, as $R \neq 0$. Therefore the required ratio is $3: 8$.