13.3.25 ㅊ On a piece of paper, there is a circle $O$ with radius $R$ and a fixed point $A$ inside the circle, and $O A=a$. Fold the paper so that a certain point $A^{\prime}$ on the circle exactly coincides with point $A$. Each such fold leaves a straight crease. When $A^{\prime}$ takes all points on the circumference, find the set of points on all the crease lines.
As shown in the figure, with $O$ as the origin and the line $OA$ as the $x$-axis, we establish a Cartesian coordinate system, so the coordinates of point $A$ are $(a, 0)$. Before folding, point $A^{\prime}(R \cos \alpha, R \sin \alpha)$ on $\odot O$ coincides with point $A$, and the crease is the line $MN$, which is the perpendicular bisector of segment $AA^{\prime}$. Let $P(x, y)$ be any point on $MN$, then $\left|P A^{\prime}\right|=|P A|$, thus
$$
(x-R \cos \alpha)^{2}+(y-R \sin \alpha)^{2}=(x-a)^{2}+y^{2},
$$

which simplifies to $2 R(x \cos \alpha+y \sin \alpha)=R^{2}-a^{2}+2 a x$, so
$$
\frac{x \cos \alpha+y \sin \alpha}{\sqrt{x^{2}+y^{2}}}=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}},
$$

leading to $\sin (\theta+\alpha)=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}}$, where $\sin \theta=$ $\frac{x}{\sqrt{x^{2}+y^{2}}}, \cos \theta=\frac{y}{\sqrt{x^{2}+y^{2}}}$. Therefore, $\left|\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}}\right| \leqslant 1$. Squaring this, we get
$$
\frac{\left(x-\frac{a}{2}\right)^{2}}{\left(\frac{R}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{R}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}} \geqslant 1,
$$

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