1. Let $A$ and $B$ be two given points, and $m$ be a given positive number. Find the geometric locus of points $M$ such that $\overline{A M}=m \cdot \overline{B M}$.
Solution. Let $O$ be the point on the segment $AB$ for which $\overline{A O}: \overline{O B}=m$. If we set $\overline{O B}=a$ and if we place a rectangular coordinate system with the origin at point $O$ and the abscissa axis along the line $AB$, then we will have $A(-m a, 0)$ and $B(a, 0)$. If $M(x, y)$ is a point of the desired geometric locus, from the condition $\overline{A M}=m \cdot \overline{B M}$ we get

$$
(x+m a)^{2}+y^{2}=m^{2}\left[(a-x)^{2}+y^{2}\right]
$$

which, after rearrangement, transforms into the form

$$
(m-1) x^{2}+(m-1) y^{2}=2 m a x
$$

The last equation defines the geometric locus of points and from it we see that for $m=1$ the geometric locus is the line $x=0$, while for $m \neq 1$ the geometric locus is the circle

$$
\left(x-\frac{m a}{m-1}\right)^{2}+y^{2}=\frac{m^{2} a^{2}}{(m-1)^{2}}
$$