5. A semicircle of radius 1 is drawn inside a semicircle of radius 2, as shown in the diagram, where $O A=O B=2$.
A circle is drawn so that it touches each of the semicircles and their common diameter, as shown.
What is the radius of the circle?
SOLUTION
CoMMENTARY
A tangent to a circle is perpendicular to the radius through the point of contact; in particular, it follows that the line of centres of two tangent circles passes through the point of tangency. We make use of these facts several times in what follows; for example, when we mention collinearity.

Let the centre of the small semicircle be $C$, that of the circle be $P$, and let the circle touch the small semicircle at $T$, noting that $C, T$ and $P$ are collinear.

Let the circle touch the large semicircle at $S$. Then $O, P$ and $S$ are also collinear.

Finally let the circle touch the diameter of the large semicircle at $R$, so that $\angle P R C=90^{\circ}$, as shown.
We have $O A=2$ and $O C=1$. Let the radius of the circle be $r$, so that $P O=2-r$, and let $O R=x$.
Now, applying Pythagoras' theorem to triangle $O R P$, we get $x^{2}+r^{2}=(2-r)^{2}$. It follows that
$$
x^{2}=4-4 r \text {. }
$$

Also, applying Pythagoras' theorem to triangle $C R P$, we have $(1+x)^{2}+r^{2}=(1+r)^{2}$. It follows that
$$
2 x+x^{2}=2 r \text {. }
$$

Hence, from equations (1) and (2), $x^{2}=4-2\left(2 x+x^{2}\right)$ and so $3 x^{2}+4 x-4=0$, that is, $(3 x-2)(x+2)=0$. Taking the positive root (the negative root being impracticable here), we have $x=\frac{2}{3}$ and so $r=\frac{8}{9}$.
Thus the radius of the circle is $\frac{8}{9}$.