Example 3 A paper has a circle $O$ with radius $R$ and a fixed point $A$ inside the circle, and $O A=a$. Fold the paper so that a point $A^{\prime}$ on the circumference coincides exactly with $A$, thus each fold leaves a straight line crease. When $A^{\prime}$ covers all points on the circumference, find the set of points on all the crease lines.
Analysis: This problem essentially involves the principle of a folded ellipse.
Solution: Establish a Cartesian coordinate system. Let the center of the circle $O$ be the origin, and the line $OA$ be the $x$-axis, with $A(a, 0)$ where $a>0$. Let any point on the circle be $A'(R \cos \alpha, R \sin \alpha)$, and let $MN$ be the perpendicular bisector of $AA'$. Suppose $P(x, y)$ is any point on $MN$, then
$$
|PA|^2 = |PA'|^2.
$$

This means
$$
(x-a)^2 + y^2 = (x - R \cos \alpha)^2 + (y - R \sin \alpha)^2.
$$

This simplifies to $2R(x \cos \alpha + y \sin \alpha) = R^2 - a^2 + 2ax$, and $\sin(\theta + \alpha) = \frac{R^2 - a^2 + 2ax}{2R \sqrt{x^2 + y^2}}$, where $\sin \theta = \frac{x}{\sqrt{x^2 + y^2}}$ and $\cos \theta = \frac{y}{\sqrt{x^2 + y^2}}$.
Thus, $\left|\frac{R^2 - a^2 + 2ax}{2R \sqrt{x^2 + y^2}}\right| \leq 1$. Squaring and rearranging gives:
$$
\frac{\left(x - \frac{a}{2}\right)^2}{\left(\frac{R}{2}\right)^2} + \frac{y^2}{\left(\frac{R}{2}\right)^2 - \left(\frac{a}{2}\right)^2} \geq 1.
$$

Therefore, the set of points is the exterior (including the boundary) of the ellipse $\frac{\left(x - \frac{a}{2}\right)^2}{\left(\frac{R}{2}\right)^2} + \frac{y^2}{\left(\frac{R}{2}\right)^2 - \left(\frac{a}{2}\right)^2} = 1$.