15. On a piece of paper, there is a circle $O$ 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 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.
$\frac{\left(x-\frac{a}{2}\right)^{2}}{\left(\frac{R}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{R}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}}=1$ outside (including the boundary) part
15. 【Analysis and Solution】As shown in the figure, with $O$ as the origin and the line $OA$ as the $x$-axis, we establish a Cartesian coordinate system, then we have $A(a, 0)$. Suppose when folding, point $A^{\prime}(R \cos \alpha, R \sin \alpha)$ on $\odot O$ coincides with point $A$, and the fold line is the line $M N$, then $M N$ is the perpendicular bisector of segment $A A^{\prime}$. Let $P(x, y)$ be any point on $M N$, then $\left|P A^{\prime}\right|=|P A|$.
$$
\begin{array}{l}
\therefore(x-R \cos \alpha)^{2}+(y-R \sin \alpha)^{2} \\
\quad=(x-a)^{2}+y^{2},
\end{array}
$$

i.e., $2 R(x \cos \alpha+y \sin \alpha)=R^{2}-a^{2}+2 a x$.
$$
\therefore \frac{x \cos \alpha+y \sin \alpha}{\sqrt{x^{2}+y^{2}}}=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}} \text {, }
$$

we get $\sin (\theta+\alpha)=\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}}$, where
$$
\begin{array}{l}
\sin \theta=\frac{x}{\sqrt{x^{2}+y^{2}}}, \cos \theta=\frac{y}{\sqrt{x^{2}+y^{2}}}, \\
\therefore\left|\frac{R^{2}-a^{2}+2 a x}{2 R \sqrt{x^{2}+y^{2}}}\right| \leqslant 1 .
\end{array}
$$
(This inequality can also be directly obtained by the Cauchy-Schwarz inequality) Squaring it, we get
$$
\frac{\left(x-\frac{a}{2}\right)^{2}}{\left(\frac{R}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{R}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}} \geqslant 1,
$$

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