4. In space, there are two intersecting perpendicular lines. Find the geometric locus of the midpoints of segments of a given length $d$, the ends of which lie on these lines.
4. Let's construct a plane $\sigma$, parallel to the given lines $l$ and $m$ and equidistant from them (Fig. 35). Consider the projections $l_{1}$ and $m_{1}$ of the given lines $l$ and $m$ onto the plane $\sigma$ and the projection $C D$ of one of the considered segments $A B$ of a given length $d$. We have: $A C = B D$ and $A C \| B D$. Therefore, $A D B C$ is a parallelogram. The midpoint $M$ of the segment $A B$ will also be the midpoint of the diagonal $C D$. In the right triangle $C A M$, the length of the hypotenuse $A M$ is $\frac{d}{2}$, and $A C = \frac{h}{2}$, where $h$ is the length of the common perpendicular to the lines $m$ and $l$. Therefore, the length of $C M$ is $\frac{1}{2} \sqrt{d^{2}-h^{2}}$, i.e., it does not depend on the position of the segment $A B$. Consequently, the segment $C D$ also maintains a constant length. Thus, we need to find in the plane the geometric locus of the midpoints of the segment $C D$ of length $\sqrt{d^{2}-h^{2}}$, whose ends slide along a pair of perpendicular lines $l_{1}$ and $m_{1}$. But in the right triangle $C O D$ (where $O$ is the point of intersection of the lines $l_{1}$ and $m_{1}$), the median $M O$ is equal to half the hypotenuse $C D$ and, therefore, the length of the median is constant, i.e., the desired geometric locus is a circle in the plane $\sigma$ with center at point $O$ and radius $\frac{1}{2} \sqrt{d^{2}-h^{2}}$.

By constructing the figure in reverse order, we can verify that any point on the considered circle is the midpoint of some segment $A B$ of length $d$, whose ends lie on the lines $l$ and $m$.

![](https://cdn.mathpix.com/cropped/2024_05_21_d3653343768e91d6de9ag-052.jpg?height=528&width=850&top_left_y=495&top_left_x=174)

Fig. 35.