In the plane, there are 3 points: $A, B, C$, and a distance $d$. Construct a circle such that the length of the tangents drawn from the given points to the circle is $d$. What is the solvability condition?
Consider the problem as solved. Let the points of tangency of the tangents of length $d$ on the sought circle be $E_{1}, E_{2}, E_{3}$ (see the figure - of the 2-2 symmetric tangents, for simplicity, only one is shown).

![](https://cdn.mathpix.com/cropped/2024_05_02_b138f9946706aa31dd51g-1.jpg?height=673&width=683&top_left_y=230&top_left_x=710)

Denoting the center of the resulting circle by $O$, the triangles $A E_{1} O$, $B E_{2} O$, and $C E_{3} O$ are congruent right triangles, because one of their legs is $d$, the other leg is the radius $r^{\prime}$ of the resulting circle, and thus their hypotenuses are equal, i.e., $O A = O B = O C = r$ is the radius of the circumcircle of the triangle defined by the points $A, B, C$, and $O$ is the center of this circumcircle. The sought circle is therefore concentric with the circumcircle of $A B C$ and its radius is $r^{\prime} = \sqrt{r^{2} - d^{2}}$.

Given $d$, the construction of $O$ (and $r$) is known, and the construction of $r^{\prime}$ is trivial.

The conditions for solvability are:

1. $A, B, C$ should not lie on a straight line,
2. $d < r$.

If $d = r$, then $r^{\prime} = 0$, meaning the resulting circle degenerates into a point.

If these conditions are met, there is always one and only one solution.

Janky Béla (Miskolc VIII. energy industry technical I. o. t.)