Example 4
Let the point set $M$ consist of two non-overlapping circles $\odot\left(O_{1}, r_{1}\right)$ and $\odot\left(O_{2}, r_{2}\right)$ (in set notation, this can be written as $M=\odot\left(O_{1}, r_{1}\right) \cup \odot\left(O_{2}, r_{2}\right)$). Find the covering radius of $M$.
Let the line $O_{1} O_{2}$ intersect $\odot O_{1}$ at $B_{1}, C_{1}$, and intersect $\odot O_{2}$ at $B_{2}, C_{2}$, and $O_{1}, C_{1}, B_{2}, O_{2}$ are all on the segment $B_{1} C_{2}$ (Figure 1.4). Construct $\odot O$ with $B_{1} C_{2}$ as its diameter. We will prove that this circle is the smallest covering circle, with the covering radius $r=\frac{1}{2} B_{1} C_{2}$.

For any point $A_{1}$ in $\odot O_{1}$, connect $A_{1} B_{1}, A_{1} C_{1}, A_{1} C_{2}$. Since $B_{1} C_{1}$ is the diameter of $\odot O_{1}$, we have $\angle B_{1} A_{1} C_{1} \geqslant 90^{\circ}$, thus
$$
\angle B_{1} A_{1} C_{2} \geqslant \angle B_{1} A_{1} C_{1} \geqslant 90^{\circ},
$$
so $A_{1}$ is inside $\odot O$. Therefore, $\odot O$ covers $\odot O_{1}$. Similarly, $\odot O$ covers $\odot O_{2}$, so $\odot O$ covers $M$.

Any circle that covers $M$ must at least cover the points $B_{1}, C_{2}$, hence its diameter $\geqslant B_{1} C_{2}$. Therefore, $\odot O$ is the smallest covering circle.

It is easy to see that when $\odot O_{1}$ and $\odot O_{2}$ are externally separated, externally tangent, or intersecting, $r$ is greater than, equal to, or less than $r_{1}+r_{2}$.
We can also consider using triangles or polygons to cover the circles.