5. As shown in Figure 3, the radii of the three semicircles are all $R$, and their centers $C_{1}, C_{2}, C_{3}$ are on the same straight line, with each center lying on the circumference of the other semicircle. $\odot C_{4}$ is tangent to these three semicircles. Let $r$ represent the radius of $\odot C_{4}$. Then, $R: r$ equals ( ).
(A) $15: 4$
(B) $11: 3$
(C) $4: 1$
(D) $3: 1$
5.C.

As shown in Figure 7, connect $C_{1} C_{4}$ and $C_{2} C_{4}$. Then $C_{2} C_{4}=R-r$,
$$
\begin{array}{r}
C_{1} C_{4}=R+r, C_{1} C_{2}=R . \text { Therefore, } \\
(R-r)^{2}+R^{2}=(R+r)^{2} .
\end{array}
$$

Solving this, we get $R^{2}=4R$.
Thus, $R: r=4: 1$.