Let $AB$ and $CD$ be the diameters of a unit circle centered at $O$, perpendicular to each other. Construct the line passing through point $A$ that cuts a segment of length $d$ from the circular segment $BOC$.
Consider the task as solved! Let the sought line intersect the radius $O C$ at point $M$, and the arc $B C$ at point $E$! Let $A M = x$ (Figure 1).

# 1986-09-260-1.eps 

Figure 1

If we can construct the segment of length $x$, then we have essentially solved the problem, since the circle centered at $A$ with radius $x$ will intersect the diameter $C D$ at point $M$, and $A M$ will be the desired line.

The triangles $A B E$ and $A M O$ are similar because their angles are equal. For the ratio of their corresponding sides, we get:

$$
\frac{A E}{A B} = \frac{A O}{A M} \quad \text{that is: } \frac{x + d}{2} = \frac{1}{x}
$$

Rearranging:

$$
x^{2} + d x - 2 = 0
$$

We are only interested in the positive solution of this equation, so:

$$
x = \frac{-d + \sqrt{d^{2} + 8}}{2}
$$

Then, given $d$, the segment $\sqrt{d^{2} + 8}$, and thus $x$ can be easily constructed (Figure 2).

$$
\text{1986-09-261-1.eps}
$$

$$
\text{Figure 2}
$$

Since the above steps are reversible, the point $M$ constructed with the segment $x$ will indeed satisfy $M E = d$, so we do get the solution.

The problem clearly has a solution only if

$$
0 \leq d \leq 1
$$

Then a simple calculation shows that

$$
1 \leq \frac{-d + \sqrt{d^{2} + 8}}{2} \leq \sqrt{2}
$$

Thus, the circle centered at $A$ with radius $x$ intersects - or touches at $O$ if $d = 1$ - the segment $O C$, meaning that if (1) is satisfied, the problem has a solution, and indeed only one solution.