We can calculate the line segment lengths as $x_k=l_{k+1}=\sqrt{1+(r^2-1)\cos^2(\frac{2\pi k}{n})}-\sqrt{r^2-1}\cos(\frac{2\pi k}{n})$ where $k=0,1,2,...,n-1$.
For $n=4m+1$, I wrote the product in the form:
$$\prod_{k=0}^{n-1}x_k=x_{\frac{n-1}{4}}\prod_{k=0}^{\frac{n-5}{4}}x_{k}x_{k+\frac{n+1}{2}}\prod_{k=\frac{n+3}{4}}^{\frac{n-1}{2}}x_{k}x_{k+\frac{n-1}{2}}.$$
Notice that $x_{\frac{n-1}{4}}\rightarrow 1$ as $n\rightarrow\infty$.
Now, it is enough to show that
$$x_{k+\frac{n\pm 1}{2}}=x_{k+\frac{n}{2}}+O(\frac{1}{n^2}),\tag{1}$$
since then $\prod_{k=0}^{n-1}x_k=(1+O(\frac{1}{n^2}))^{\frac{n-1}{2}}$ and when $n\rightarrow\infty$, it tends to $1$.
We can show that
$$\cos(\frac{2\pi}{n}(k+\frac{n\pm 1}{2}))=-\cos(\frac{2\pi k}{n})+O(\frac{1}{n^2})$$
and since also $\sqrt{1+\frac{1}{n^2}}=1+O(\frac{1}{n^2})$, $(1)$ holds.
$n=4m+3$ case is similar.
Let us collect the operations (Ops) that cyclos (or John if you prefer) can do besides the given operations. Call a line a drawn line if there are at least two mark points on it. We also say cyclos can "draw" that line.
Op foot: Suppose $a$ is a marked point and $\ell$ is a drawn line. Then cyclos can mark the foot of perpendicular from $a$ to $\ell$.
Op perpendicular: Suppose $a$ is a marked point and $\ell$ is a drawn line ($a$ might be on $\ell$). Then cyclos can draw the line through $a$ that is perpendicular to $\ell$.
Op far: Suppose $a,b$ are two marked points. Then cyclos can mark a point that is far away from line $\ell$ that passes $a,b$. All we need is that the distance between the new point and $\ell$ is at least $\|\overline{ab}\|$, which is the distance between $a$ and $b$.
Op midpoint: Suppose $a,b$ are two marked points. Then cyclos can mark the midpoint of $a,b$.
Op $\frac14$ away: Suppose $a,b$ are two marked points. Then cyclos can mark a point that is $\frac14\|\overline{ab}\|$ away from line $ab$.
Op rotate $90°$: Suppose $a,b$ are two marked points. Then cyclos can mark a point that together with $a,b$ form an isosceles right triangle with $b$ at the right angle.
Op double: Suppose $a,b$ are two marked points. Then cyclos can mark the point $c$ such that $b$ is the midpoint of $a$ and $c$.
Op compass: Suppose $a,b$ are two marked points. Then cyclos can draw a circle that is centered $b$ and passing $a$. This is the operation that we are asked to show.
Here is how cyclos can do the operations above.
Op foot: Let $b,c$ are two marked points on $\ell$. Draw the circle with diameter $\overline{ab}$ and circle with diameter $\overline{ac}$. Mark the intersection of the two circles, which is the foot of perpendicular from $a$ to $\ell$.
Op perpendicular: Let $b,c$ are two marked points on $\ell$. Draw the circle with diameter $\overline{bc}$. Mark a point $d$ on the circle that is neither $b$ nor $c$. Mark point $e$, the foot of perpendicular from $d$ to $\ell$.
- If $e$ is $a$, line $ad$ is the wanted line.
- Otherwise draw a similar line $\overline{fg}$ that is perpendicular to $\ell$. Mark $h$, the foot of perpendicular from $d$ to $\overline{fg}$. Mark $i$, the foot of perpendicular from $a$ to line $\overline{dh}$. The line $\overline{ai}$ is what we wanted.
Op far: this is obvious and probably not interesting to most readers, so I will omit how cyclos can achieve it, although it take a while to show a random algorithm with high probablity and it takes much longer to show a deterministic one.
Op midpoint: Mark $c$, a point that is far away from line $\overline{ab}$. Draw line $\ell_a$ and $\ell_b$ that are perpendicular to $\overline{ab}$, passing through $a$ and $b$ respectively. Mark the foot of perpendicular from $c$ to $\ell_a$ and to $\ell_b$, namely $d$ and $e$ respectively. Draw the circles with diameter $\overline{ad}$ and $\overline{be}$ respectively. Mark $e$, one of the intersection point of two circles. Mark the foot of perpendicular from $e$ to line $\overline{ab}$. $e$ is the midpoint of $\overline{ab}$.
Op $\frac14$ away: Mark $c$, the midpoint of $a$ and $b$. Mark the midpoint of $a$ and $c$ as well as the midpoint of $c$ and $b$. Repeat until all points on line segment $\overline{ab}$ that are $\frac i{16}$ away from $a$ are marked, assuming WLOG the length of line segment $\overline{ab}$ is $1$. Draw the circle with diameter that ends at $a$ and the marked point that is $\frac{10}{16}$ away from $a$. Draw the circle with diameter that ends at $b$ and the marked point that is $\frac{10}{16}$ away from $b$. Mark $x$, one of the intersection points of two circles. Then $x$ is a wanted point.
Op rotate $90°$**: Mark $x$, a point that is $\frac14\|\overline{ab}\|$ away from line $ab$. Mark $d$, the foot of perpendicular from $x$ to the line that is through $b$ and perpendicular to line $ab$. Note that we just rotated $\frac14$ of line segment $\|\overline{ab}\|$ $90°$ to $\overline{bd}$.
Mark $c$, the foot of perpendicular from $d$ to the line that is through $a$ and perpendicular to line $ab$. Then $abdc$ is a rectangle. We can, similarly, rotate $\frac14$ of line segment $\|\overline{cd}\|$ to $\overline{df}$. Repeat two more times to get $\overline{fh}$ and $\overline{hj}$. The $j$ is the wanted point.
Op double: Apply op rotate $90°$ two times.
Op compass: Apply op double to mark point $c$. Draw the circle with diameter $\overline{ac}$. This is the wanted circle.
(Graphs will come soon.)
Best Answer
No that's not true for $n=3$. Consider 7 points on the complex plane in such location: one at $0$ and other 6 at $0.01\omega^i,\omega^i, i=0,1,2$, where $\omega$ is the 3-th unit root ($-\frac{1}{2}+\frac{\sqrt{3}}{2}i$).