Example 6 (1996 National High School League Question) A frustum of height 8 contains a sphere $O_{1}$ with a radius of 2, the center of which, $O_{1}$, lies 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.
Solution: Choose B. Reason: Perform dimensional reduction, make an axial section of the frustum through $O_{2}$, as shown in Figure 2-1(1), and then make a section through $O_{2}$ perpendicular to the axis of the frustum, intersecting the axis at circle $O$, as shown in Figure 2-1(2).
From the figure, 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 so that these points and the circle with center $O_{2}$ and radius 3 have at most one common point. From Figure 2-1(1), we know $\sin 45^{\circ}<\sin \theta=\frac{3}{4}<\sin 60^{\circ}$, thus $45^{\circ}<\theta<60^{\circ}$, so the maximum number of 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.