1. Given are the circle $(x-5)^{2}+y^{2}=9$ and the line $x+2=0$. Find the geometric place of points equally distant from the circle and the line.
Solution. Let $k$ and $p$ be the given circle and line. Let $M(X, Y)$ be a point in the plane whose distances to $p$ and $k$ are $d_{1}$ and $d_{2}$ respectively (see diagram), where

$$
\begin{aligned}
& d_{1}=|X+2| \text { and } \\
& d_{2}=|\overline{M C}-3|=\left|\sqrt{(X-5)^{2}+Y^{2}}-3\right| .
\end{aligned}
$$

The point $M$ belongs to the desired geometric locus if and only if $d_{1}=d_{2}$, i.e.,

$$
|X+2|=\left|\sqrt{(X-5)^{2}+Y^{2}}-3\right|
$$

Furthermore,

$$
X+2>0 \text { and } \sqrt{(X-5)^{2}+Y^{2}}-3>0
$$

(see diagram). Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_83e7cab9d3c7aa2efae6g-6.jpg?height=423&width=589&top_left_y=690&top_left_x=859)

$$
X+2=\sqrt{(X-5)^{2}+Y^{2}}-3
$$

from which it follows that $Y^{2}=20 X$. Therefore, the desired geometric locus is the parabola $y^{2}=20 x$.