Concyclic tangent property of circles Suppose you have some $n$ points lying on the circles (n can be as large or as small as you want) spaced however you want. Now, let's say we give each point a constant velocity tangent to the circle, with the direction given by $\vec{v}= \vec{\omega} \times \vec{r}$ where $\vec{\omega}$ is a vector perpendicular to plane(you can fix it's direction to be above or below, but keep this direction same for all points) and $\vec{r}$ is a vector connecting to that point from origin of circle. Now, after some time $\Delta t$ prove that these points are still concylic.
This question arose from q.31 in page-13 in Jaan Kalda's olympiad physics notes, my personal try I tried checking the result by seeing if it was true for a shape inscribed in the circle and it looks to me that the result is more or less correct:

The black is the initial rectangle configuration made by points on the circle, green is the rectangle figure formed by points after moving with tangent velocity with some time after that blue and then purple are the figures.
However, I have pretty much no idea after this on how to prove this. Any help will be appreciated.

WLOG, let the circle be centered at origin with unit radius. Then:
Let
$$
\begin{align*}
\theta_n \in [0, 2 \pi) &\Rightarrow z_n = e^{i\theta_n} \\ 
&\Rightarrow z_n^t = z_n + (v\Delta t)e^{i(\frac{\pi}{2} + \theta_n)} = e^{i\theta_n} + (v\Delta t)e^{i(\frac{\pi}{2} + \theta_n)} \\
&\Rightarrow ||z_n^t||^2 = 1 + v^2 (\Delta t)^2
\end{align*}
$$
Qed. Also note that the new circle is still centred at the same point with ratio of radii as $\frac{1}{\sqrt{1 + v^2 (\Delta t)^2}}$.
Edit. After writing my overkill solution, I just realized that you can use Pythagorous theorem to solve it immediately.
