Given a circle $k$, a diameter of it $P Q$, and two lines $e$ and $f$ perpendicular to $P Q$. Let one of the intersection points (if any) of $k$ and $e$ be denoted by $E_{1}$, and the intersection points (if any) of the circle centered at $P$ passing through $E_{1}$ with $f$ be denoted by $F_{1}$ and $F_{2}$. What is the geometric locus of the points $F_{1}, F_{2}$ if $e$ and $f$ take on all positions where their distance is equal to a given segment $d$?
Let's place our figure in the coordinate system such that the center of $k$ is at the origin, its radius is the unit length, and let $P(-1 ; 0)$, then the equation of $k$ is $x^{2}+y^{2}=1$. Furthermore, let the equation of the line $e$ be $x=e$, and in the scenarios we are examining, $-1 \leq e \leq 1$, so the equation of $f$ is $x=e \mp d$ depending on whether it is to the left or right of $e$.
![](https://cdn.mathpix.com/cropped/2024_05_02_bdbd2ece8648ba203f40g-1.jpg?height=656&width=1506&top_left_y=314&top_left_x=298)

The coordinates of $E_{1}$ are $\left(e, \sqrt{1-e^{2}}\right)$, indeed, based on symmetry, it is sufficient to take the one with a non-negative ordinate. Similarly, $F_{1}$ and $F_{2}$ are points symmetric with respect to the $x$-axis, so the $x$-axis will also be the axis of symmetry for the geometric locus we are looking for. The radius of the circle centered at $P$ and passing through $E_{1}$ is:

$$
P E_{1}^{2}=(e+1)^{2}+\left(1-e^{2}\right)=2+2 e
$$

Thus, the equation of this circle is: $(x+1)^{2}+y^{2}=2+2 e(\geq 0$.

From this, we get the square of the $\pm y$ ordinate of the two $F$ points by substituting $x=e \mp d$:

$$
y^{2}=2+2 e-(e \mp d+1)^{2}=-(e \mp d)^{2}+1 \pm 2 d
$$

Therefore, the sum of the squares of the coordinates of $F_{1}$ and $F_{2}$ does not depend on the position of $e$, but only on the distance between $e$ and $f$ - which is constant - and is

$$
x^{2}+y^{2}=1 \pm 2 d
$$

Accordingly, the points $F$ lie on one or the other of the circles $k^{\prime}$ and $k^{\prime \prime}$, which are concentric with $k$, with radii $\sqrt{1+2 d}$ (> 1), when $f$ follows $e$ from the left, and $\sqrt{1-2 d}$ (1/2 in this case), when $f$ is to the right of $e$, in which case there are no $F_{1,2}$ points.

We need to examine whether every point of $k^{\prime}$ and $k^{\prime \prime}$ belongs to the geometric locus. $k^{\prime}$ extends beyond $k$ in both directions along the $x$-axis, so only a part of it can meet the criteria. When $e$ is at its rightmost position (giving $E_{1}$), $e=1$, so only the arc of $k^{\prime}$ where $x \leq 1-d$ can be considered. Every point on this arc belongs to the geometric locus, because even the leftmost point $F^{*}$ of $k^{\prime}$ can be traced back to the point $E^{*}$ on $k$ that generates it, indeed

$$
F^{*} P=\sqrt{1+2 d}-14$.)

If, however, $f$ is to the right of $e$, then $k^{\prime \prime}$ is inside $k$ (for $d<1 / 2$), and every point of it belongs to the geometric locus. (On the other hand, not every point of $k$ generates an $F$; let's denote the leftmost and rightmost points of $k^{\prime \prime}$ by $B$ and $J$, respectively, so only those points generate an $F$ for which $P B \leq P E_{1} \leq P J$.)