2. Given a circle $(O, r)$, its radius $A B$ and a point $M$ on the circle. Let $Q$ be the projection of $M$ on $A B$, $N$ be the midpoint of $M Q$, and $D$ be the other intersection point of the line $A N$ and the given circle. Determine the geometric locus of the intersection $P$ of the lines $B D$ and $Q M$, as the point $M$ moves along the circle $(O, r)$.
Solution. Let us choose a coordinate system such that point $O$ is the origin, and points $A$ and $B$ lie on the $x$-axis with coordinates $(-r, 0)$ and $(r, 0)$, respectively, as shown in the diagram on the right. Let point $M$ have coordinates $\left(x_{0}, y_{0}\right)$, where $x_{0}^{2}+y_{0}^{2}=r^{2}$ and $x_{0} \neq \pm r$. Then point $N$ has coordinates $\left(x_{0}, \frac{y_{0}}{2}\right)$. Since

$$
\measuredangle Q A N=\measuredangle B A D=90^{\circ}-\measuredangle D B A=90^{\circ}-\measuredangle P B D=\measuredangle Q P B
$$

we conclude that triangles $A N Q$ and $P B Q$ are similar,

![](https://cdn.mathpix.com/cropped/2024_06_05_c53e313ed51c88c0c5c5g-4.jpg?height=445&width=281&top_left_y=268&top_left_x=1181)
so

$$
P Q=\frac{A Q \cdot B Q}{N Q}=\frac{\left(r+x_{0}\right)\left(r-x_{0}\right)}{\frac{y_{0}}{2}}=2 y_{0}
$$

Therefore, point $P$ has coordinates $(x_{0}, 2 y_{0})$, and thus lies on the ellipse

$$
\frac{x^{2}}{r^{2}}+\frac{y^{2}}{4 r^{2}}=1
$$

It is easily verified that every point on this ellipse, except for points $A(-r, 0)$ and $B(r, 0)$, can be obtained in the described manner.