Example 9 (1996 National High School League Question) A frustum of height 8 contains a sphere $O_{1}$ with a radius of 2, the center $O_{1}$ is on the axis of the frustum, and the sphere $O_{1}$ is tangent to the top base and the side of the frustum. Another sphere $O_{2}$ with a radius of 3 can be placed inside the frustum, such that the sphere $O_{2}$ is tangent to the sphere $O_{1}$, the bottom base, and the side of the frustum at only one point. Apart from the sphere $O_{2}$, the maximum number of additional spheres with a radius of 3 that can be placed inside the frustum is ( ).
A. 1
B. 2
C. 3
D. 4
Solution: Choose B. Reason: Make the axial section of the frustum passing through $O_{2}$, as shown in Figure 7-8(1). Then make a section through $O_{2}$ perpendicular to the axis of the frustum, intersecting the axis at circle $O$. From Figure 7-8(1), it is easy to find that $O O_{2}=4$.

This problem is equivalent to: On a circle with center $O$ and radius 4, except for $O_{2}$, how many more points can be placed such that the circles with these points and $O_{2}$ as centers and 3 as the radius have at most one common point.

From Figure 7-8(2), $\sin 45^{\circ}<\sin \theta=\frac{3}{4}<\sin 60^{\circ}$, then $45^{\circ}<\theta<60^{\circ}$, so the maximum number of additional points that can be placed is $\left[\frac{360^{\circ}}{2 \theta}\right]-1=3-1=2$, satisfying the above requirements, i.e., the frustum can contain at most 2 more spheres with a radius of 3.