## Task $2 / 62$

It is to determine the geometric locus of all points where the ratio

$$
\frac{P P_{1}}{P P_{2}}
$$

of the distances $P P_{1}$ and $P P_{2}$ to two given points $P_{1}$ and $P_{2}$ has the constant value $c$.
Let $P_{1}$ and $P_{2}$ be the given points, $a$ their distance. The coordinate system (right-angled Cartesian) is conveniently set such that $P_{1}$ is at the origin and $P_{2}$ lies on the positive x-axis. Then, for any point $P(x ; y)$ on the geometric locus (illustration):

$$
x^{2}+y^{2}=u^{2} \quad ; \quad(x-a)^{2}+y^{2}=v^{2}
$$

If the first equation is divided by the second, we obtain

$$
\frac{x^{2}+y^{2}}{(x-a)^{2}+y^{2}}=\frac{u^{2}}{v^{2}}=c^{2}
$$

From this, it follows that $x^{2}+y^{2}=c^{2}(x-a)^{2}+c^{2} y^{2}$, and by rearranging,

$$
x^{2}-2 x \frac{c^{2} a}{c^{2}-1}+y^{2}+\frac{c^{2} a^{2}}{c^{2}-1}=0 \rightarrow\left(x-\frac{a c^{2}}{c^{2}-1}\right)^{2}+y^{2}=\left(\frac{a c}{c^{2}-1}\right)^{2}
$$

It is evident that the geometric locus is a circle with the center coordinates $M\left(\frac{a c^{2}}{c^{2}-1} ; 0\right)$ and the radius $r=\frac{a c}{c^{2}-1}$.