Example 1 A paper has a circle $\odot O$ with radius $R$ and a fixed point $A$ inside the circle, where $O A=a$. Fold the paper so that a point $A^{\prime}$ on the circumference coincides exactly with point $A$. Each such fold leaves a straight line crease. Find the set of points on all the crease lines when $A^{\prime}$ takes all points on the circumference.
Solution: Establish a rectangular coordinate system as shown in Figure 2. By the symmetry of the paper folding, point $A^{\prime}$ and point $A$ are symmetric with respect to the line $l$ of the fold, i.e., $l$ is the perpendicular bisector of segment $A A^{\prime}$.
Connecting $O A^{\prime}$ and intersecting $l$ at point $P$, we have
$$
P O+P A=P O+P A^{\prime}=O A^{\prime}=R \text {. }
$$

Therefore, point $P$ lies on the ellipse with foci at $O$ and $A$ and a major axis length of $R$.

Let $P^{\prime}$ be any point on the line $l$ different from point $P$. Then
$$
P^{\prime} O+P^{\prime} A=P^{\prime} O+P^{\prime} A^{\prime}>O A^{\prime}=R .
$$

Thus, point $P^{\prime}$ lies outside the aforementioned ellipse.
Hence, the set of points on the line $l$ of the fold is on or outside the ellipse with foci at $O$ and $A$ and a major axis length of $R$. The equation of the ellipse is
$$
\frac{\left(x-\frac{a}{2}\right)^{2}}{\frac{R^{2}}{4}}+\frac{y^{2}}{\frac{R^{2}}{4}-\frac{a^{2}}{4}} \geqslant 1 .
$$