Given a circle $k$ and two points $A$ and $B$ outside the circle. Construct a point $C$ on the circle such that if $CA$ intersects $k$ at $D$, and $CB$ intersects $k$ at $E$, then the length of $DE$ is a given length $(l)$!
A chord $DE$ of length $l$ in circle $k$ is visible from a point $C$ on the circumference of the circle at an angle $\gamma$. (For example, in circle $k$, we measure the chord of length $l$ anywhere and draw a tangent to the circle at one of its endpoints: the angle between the chord and the tangent is $\gamma$).

![](https://cdn.mathpix.com/cropped/2024_05_02_0860e02f91d47cf764efg-1.jpg?height=430&width=406&top_left_y=300&top_left_x=838)

Accordingly, the vertex $C$ of the triangle $ABC$ lies on two circles: one is the circle $k$, and the other is the circle $k'$, from whose circumference the side $AB$ is visible at the specified angle $\gamma$.

Since in circle $k$, two different inscribed angles correspond to the chord of length $l$, i.e., $\gamma$ and $180^{\circ}-\gamma$, it is also possible in triangle $ABC$ that the angle at vertex $C$ is $180^{\circ}-\gamma$, so that the vertex $C$ lies on the circle $k$ on one hand, and on the circle $k''$ on the other, which is symmetric to $k'$ with respect to $AB$.

Thus, the number of solutions is: $4,3,2,1,0$.