3.17. Form the equation of the geometric locus of points on the plane, each of which is twice as close to point $A(4 ; 0)$ as to point $B(1 ; 0)$.
Solution. The distance from the current point $M(x ; y)$ of the desired curve to the point $A(4 ; 0)$ is denoted by

$$
r_{1}=\sqrt{(x-4)^{2}+(y-0)^{2}},
$$

to the point $B(1 ; 0)$ - by

$$
r_{2}=\sqrt{(x-1)^{2}+(y-0)^{2}}
$$

From the condition of the problem, it follows that $r_{1}=2 r_{2}$, from which we obtain

$$
\sqrt{(x-4)^{2}+y^{2}}=2 \sqrt{(x-1)^{2}+y^{2}}
$$

Squaring both sides of the last relation, we get $(x-4)^{2}+y^{2}=4\left((x-1)^{2}+y^{2}\right)$. Expanding the brackets and combining like terms: $x^{2}+y^{2}=(2 \sqrt{3})^{2}$ - the equation of a circle centered at the origin with radius $2 \sqrt{3}$.