47. Let (Fig. 9) $\breve{B C}=2 \alpha, \breve{B L}=2 \beta$. Then

$$
|A C|=2 R \cos \alpha, \quad|C L|=2 R \sin (\alpha+\beta)
$$

$|C M|=|C L| \cos \left(90^{\circ}-\beta\right)=2 R \sin (\alpha+\beta) \sin \beta$,

$$
\begin{aligned}
A M \mid & =|A C \cdot|-|C M|=2 R[\cos \alpha-\sin (\alpha+\beta) \sin \beta]= \\
& =2 R\left[\cos \alpha-\frac{1}{2} \cos \alpha+\frac{1}{2} \cos (\alpha+2 \beta)\right]=2 R \cos \beta \cos (\alpha+\beta)
\end{aligned}
$$

and, $\quad$ finally, $\quad|A N|=a=|A M| \cos \alpha=2 R \cos \alpha \cos \beta \cos (\alpha+\beta)$. On the other hand, if $K, P$ and $Q$ are the midpoints of $A O, C O$ and $C L$ respectively, then $|K P|=\frac{1}{2}|A C|=R \cos \alpha,|P Q|=\frac{R}{2}, \widehat{K P Q}=$ $=\widehat{K P O}+\widehat{O P Q}=\alpha+180^{\circ}-\widehat{C O L}=\alpha+180^{\circ}-2 \alpha-2 \beta=180^{\circ}-\alpha-$ $-2 \beta$ and, by the cosine rule, $|K Q|^{2}=\frac{R^{2}}{4}+R^{2} \cos ^{2} \alpha+R^{2} \cos \alpha \times$ $x \cos (\alpha+2 \beta)=\frac{R^{2}}{4}+2 R^{2} \cos \alpha \cos \beta \cos (\alpha+\beta)=\frac{R^{2}}{4}+R a$.

Answer: $\sqrt{\frac{R^{2}}{4}+R a}$.
47. Let $A B$ be the diameter of a circle, $O$ its center, $|A B|=2 R, C$ a point on the circle, $M$ a point on $A C$. From $M$, a perpendicular $M N$ is dropped onto $A B$ and a perpendicular to $A C$ is erected, intersecting the circle at point $L$ (segment $C L$ intersects $A B$). Find the distance between the midpoint of $A O$ and the midpoint of $C L$, if $|A N|=a$.