Two unit-radius circles intersect at points $A$ and $B$. One of their common tangents touches the circles at points $E$ and $F$. What can be the radius of a circle that passes through points $E, F$, and $A$?
Solution. Let the centers of the unit circles be denoted by $O_{1}, O_{2}$, the center of the circle passing through points $E, F$, and $A$ by $O_{3}$, its radius by $r$, and let $A$ be the intersection point of the unit circles closer to the tangent $E F$. We know that $E O_{1} \perp E F$ and similarly $O_{3} A \perp E F$, so $E O_{1} \| O_{3} A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4e8d193cd3bf49f53f57g-1.jpg?height=485&width=461&top_left_y=305&top_left_x=821)

The quadrilateral $A O_{1} E O_{3}$ has a pair of parallel sides, so it is a trapezoid. Since $O_{1} E = O_{1} A = 1$, and $O_{3} E = O_{3} A = r$, the lengths of two adjacent sides of the trapezoid are equal, making it a kite. Since two opposite sides are parallel, the quadrilateral $A O_{1} E O_{3}$ is also a rhombus. Therefore, the radius of the circle passing through points $E, F, A$ is $O_{3} E = 1$ unit.

There is another solution to the problem if we are looking for the radius of the circle passing through the farther intersection point of the unit circles and the points $E, F$. Similarly to the previous steps, we can show that the radius of this circle is also 1 unit.