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 $A_{1}$ on the circumference coincides exactly with point $A$. Each such fold leaves a crease. Find the set of points on all the crease lines when $A_{1}$ covers all points on the circumference. (2003, National High School Mathematics Competition)
Solution: Establish a rectangular coordinate system as shown in Figure 1, let $P(x, y)$ be any point on the line $l$ (the crease), and $B$ be the midpoint of $A A_{1}$. Thus, we have
$$
\left\{\begin{array}{l}
O B=\frac{1}{2}\left(O A+O A_{1}\right), \\
P B=O B-O P, \\
A A_{1}=O A_{1}-O A, \\
P B \cdot A A_{1}=0 .
\end{array}\right.
$$

Then $\left[\frac{1}{2}\left(O A+O A_{1}\right)-O P\right] \cdot\left(O A_{1}-O A\right)=0$,
which means $\frac{1}{2}\left(O A_{1}^{2}-O A^{2}\right)-O P \cdot O A_{1}+O P \cdot O A=0$. Therefore, $O P \cdot O A_{1}=\frac{1}{2}\left(O A_{1}^{2}-O A^{2}\right)+O P \cdot O A$ $=\frac{1}{2}\left(R^{2}-a^{2}\right)+a x$.
By $\left|O P \cdot O A_{1}\right| \leqslant|O P| \cdot\left|O A_{1}\right|$, we get
$$
\left|\frac{1}{2}\left(R^{2}-a^{2}\right)+a x\right| \leqslant \sqrt{x^{2}+y^{2}} \cdot R \text {. }
$$

Squaring both sides, we have
$$
\begin{array}{l}
\frac{1}{4}\left(R^{2}-a^{2}\right)^{2}+a x\left(R^{2}-a^{2}\right)+a^{2} x^{2} \\
\leqslant x^{2} R^{2}+y^{2} R^{2} .
\end{array}
$$

Rearranging, we get
$$
\begin{array}{l}
\left(R^{2}-a^{2}\right)\left(x-\frac{a}{2}\right)^{2}+y^{2} R^{2} \\
\geqslant \frac{1}{4}\left(R^{2}-a^{2}\right) R^{2} .
\end{array}
$$

Since $\left(R^{2}-a^{2}\right) R^{2}>0$, we have
$$
\frac{\left(x-\frac{a}{2}\right)^{2}}{\frac{1}{4} R^{2}}+\frac{y^{2}}{\frac{1}{4}\left(R^{2}-a^{2}\right)} \geqslant 1 \text {. }
$$

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