Example (2)  
Given $\triangle ABC$, find the smallest circle that covers $\triangle ABC$.
Solution: It is evident that the circumcircle of $\triangle ABC$ covers $\triangle ABC$. At first glance, this circle seems to be the smallest circle that can cover $\triangle ABC$. However, this conclusion is not entirely correct.

We will discuss this in three cases.
(1) $\triangle ABC$ is an obtuse triangle.
Construct $\odot O$ with the longest side $BC=a$ as its diameter. For any point $A'$ belonging to $\triangle ABC$, since $\angle BA'C \geqslant \angle BAC > 90^{\circ}$, $A'$ lies inside $\odot O$, meaning $\odot O$ covers $\triangle ABC$ (Figure 1.3(1)).

Since the diameter of any circle that covers $\triangle ABC$ is not less than the diameter of $\triangle ABC$, which is $BC=a$, $\odot O$ is the smallest circle that covers $\triangle ABC$.

In this case, $\odot O$ is smaller than the circumcircle of $\triangle ABC$. Because $\angle BAC > 90^{\circ}$, $BC$ (the diameter of $\odot O$) cannot be the diameter of the circumcircle, thus it is smaller than the diameter of the circumcircle.
(2) $\triangle ABC$ is a right triangle.
Similarly to (1), the circle with the hypotenuse $BC=a$ as its diameter is the smallest circle that covers $\triangle ABC$. In this case, the smallest covering circle is the circumcircle of $\triangle ABC$ (Figure 1.3(2)).
(3) $\triangle ABC$ is an acute triangle.

In this case, the circumcircle of $\triangle ABC$ is also the smallest circle that covers $\triangle ABC$. This can be proven by showing that any other circle that covers $\triangle ABC$ is no smaller than the circumcircle of $\triangle ABC$. If $\odot O$ covers $\triangle ABC$, by appropriately moving $\triangle ABC$ within $\odot O$, we can always make two vertices, say $B$ and $C$, lie on the circumference of the circle. Specifically, we first translate $\triangle ABC$ within $\odot O$ to move one vertex $B$ to the circumference, and then rotate $\triangle ABC$ around point $B$ within $\odot O$ to move another vertex $C$ to the circumference. Since the movement is always within $\odot O$, the third vertex $A$ will lie inside or on the circumference of $\odot O$ (Figure 1.3(3)). The arc $\overparen{BC}$ (the minor arc) subtends a central angle $\delta \leqslant \angle BAC < 90^{\circ}$, so the diameter of $\odot O = \frac{BC}{\sin \delta} \geqslant \frac{BC}{\sin \angle BAC} = $ the diameter of the circumcircle of $\triangle ABC$.

Definition 4
If a point set $M$ can be covered by a circle, then we say that $M$ is a bounded point set.

Circles, triangles, polygons, and ellipses are bounded point sets, while parabolas are not bounded point sets.

Generally, a closed curve (that does not intersect itself) and its interior form a bounded point set, which we call a bounded closed set.
Definition 5
Let $M$ be a bounded point set. The smallest circle that covers $M$ is called the minimum covering circle of $M$. The radius of the minimum covering circle is called the covering radius of the point set $M$.

Let the maximum side of $\triangle ABC$ be $BC=a$, then $\angle BAC \geqslant 60^{\circ}$. Therefore, combining (1), (2), and (3) of Example 2, the covering radius $\rho$ of $\triangle ABC$ satisfies
$$
\frac{a}{2} \leqslant \rho \leqslant \frac{a}{2 \sin 60^{\circ}} = \frac{a}{\sqrt{3}},
$$

The left equality holds if and only if $\triangle ABC$ is an obtuse or right triangle, and the right equality holds if and only if $\triangle ABC$ is an equilateral triangle. The detailed derivation of the right inequality is left to the reader. This inequality provides the upper and lower bounds for the covering radius of a triangle.
The minimum covering circle and covering radius of an $n$-sided polygon will be discussed in the third section. For a regular $n$-sided polygon, the problem is straightforward and can be solved immediately.