15. On a piece of paper, there is a circle $\odot O$ with radius $R$ and a fixed point $A$ inside the circle, and $OA=a$. Fold the paper so that a point $A^{\prime}$ on the circumference of the circle coincides exactly with point $A$. Each such 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.
Solve: Establish a rectangular coordinate system as shown in Figure 4. Then we have $A(a, 0)$. Suppose when folding, point $A^{\prime}(R \cos \alpha, R \sin \alpha)$ on $\odot\left(\right.$ ) coincides with point $A$. The fold line is the line $M N$, which is the perpendicular bisector of segment $A A^{\prime}$. Let $P(x, y)$ be any point on $M N$. Then $\left|P A^{\prime}\right|=|P A|$. That is,
$$
(x-R \cos \alpha)^{2}+(y-R \sin \alpha)^{2}=(x-a)^{2}+y^{2} \text {. }
$$

Therefore,
$$
\frac{x \cos \alpha+y \sin \alpha}{\sqrt{x^{2}+y^{2}}}=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}} .
$$

We get
$$
\sin (\theta+\alpha)=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}} \text {, }
$$

Given $\quad \sin \theta=\frac{x}{\sqrt{x^{2}+y^{2}}}, \cos \theta=\frac{y}{\sqrt{x^{2}+y^{2}}}$.

Thus
$$
\left|\frac{R^{2}-a^{2}+2 a x}{2 R-x^{2}+y^{2}}\right| \leqslant 1
$$

Simplifying, we get
$$
\frac{(2 x-a)^{2}}{R^{2}}+\frac{4 y^{2}}{R^{2}-a^{2}} \geqslant 1 \text {. }
$$

Therefore, the set of points sought is $\frac{(2 x-a)^{2}}{R^{2}}+\frac{4 y^{2}}{R^{2}-a^{2}}=1$ and its exterior (including the boundary).