5. As shown in Figure 1, there are 1999 circles each tangent to both sides of $\angle M O N$, and these circles are sequentially tangent to each other. If the radii of the largest and smallest circles are 1998 and 222 respectively, then the radius of the circle in the middle is $(\quad)$.
(A) 1110
(B) 1776
(C) 666
(D) cannot be determined

保留了原文的换行和格式。
5. (C).

As shown in Figure 3. Let the radii of the externally tangent circles $\odot O_{n}, \odot O_{n+1}$ be $R_{n}$ and $R_{n+1}$, and $\angle M O N=2 \alpha$ (where $\alpha$ is a constant). Construct $O_{n} A \perp O M$ at $A$, $O_{n+1} B \perp O M$ at $B$, and $O_{n} C \perp O_{n+1} B$ at $C$.
Then $O_{n+1} C=O_{n+1} B-B C=O_{n+1} B-O_{n} A$
$$
=R_{n+1}-R_{n} \text {, }
$$
$\angle O_{n+1} O_{n} A=\angle M O O_{n+1}=\frac{1}{2} \angle M O N=\alpha$,
$$
O_{n} O_{n+1}=R_{n+1}+R_{n} \text {. }
$$

In $\mathrm{R}_{1} \triangle O_{n} O_{n+1} C$,
$$
\begin{array}{l}
\because \sin \angle O_{n+1} O_{n} A=\frac{O_{n+1} C}{O_{n} O_{n+1}}, \\
\therefore \sin \alpha=\frac{R_{n+1}-R_{n}}{R_{n+1}+R_{n}} .
\end{array}
$$

Simplifying, we get $\frac{R_{n+1}}{R_{n}}=\frac{1+\sin \alpha}{1-\sin \alpha}$ (a constant).
Thus, $\left\{R_{n}\right\}$ is a geometric sequence.
Furthermore, $R_{1}, R_{10000}, R_{1999}$ form a geometric sequence.
Therefore, $R_{10000}=\sqrt{R_{1} R_{1999}}=\sqrt{222 \times 1998}=666$.