Given a segment, and a point $A$ on the circumference of a circle. The projection of point $A$ onto the tangent line at point $P$ on the circle is $A^{\prime}$. Construct the point $P$ such that $P A^{\prime} + A^{\prime} A$ is equal to the given segment.
Imagine we have completed the construction. Measure on the line $P A^{\prime}$ from $A^{\prime}$ in the direction opposite to $P$ the distance $A A^{\prime}$.

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The point $C$ obtained satisfies $P C = d$, the given distance, and $\angle A C A^{\prime} = 45^{\circ}$, meaning that $\triangle A C A^{\prime}$ is an isosceles right triangle. The construction from here is easily readable. For any point $Q$ on the circle, construct the tangent $e$, and measure on it from $Q$ the given distance $d$. On the endpoint $R$ different from $Q$, measure the $45^{\circ}$ angle, and the ray of the angle intersects the circle at a point $M$. Then we just need to rotate the line $M R$ until $M$ is at point $A$ while ensuring that the line $R Q$ is tangent to the circle, so as $Q$ moves along the circle, it will end up at $P$.

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The $45^{\circ}$ line constructed at $R$ on $e$ either touches the circle, intersects it at two points, or does not intersect it. If it intersects at two points, then either of these points can serve as $M$, provided that it falls into the half-plane bounded by the line $O Q$ that contains $R$. Thus, we get 1 or 2 essentially different solutions. An additional solution is given by the reflection of $P$ over $A O$. Therefore, the total number of solutions is:

2, if $d < r - \sqrt{2} r$.

1, if $r - \sqrt{2} r \leq d \leq r + \sqrt{2} r$.

0, if $d > r + \sqrt{2} r$.