System of equations and recurrence relation I am trying to find the general solution for $N$ of the following system of equations
$$
\begin{cases}
    (x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\
    {x_n}^2 + {y_n}^2 = 1
\end{cases}
$$
with the initial values $x_0 = 1$ and $y_0 = 0$ and the following

*

*$\theta$ is a constant and $0 \leqslant \theta \leqslant 2$

*$N$ is a constant and we want to find the terms $(x_N, y_N)$
Using substitution with respect to $N$, we have
$$
\begin{align}
    x_0 = 1 \quad & ; \quad y_0 = 0 \\
    x_1 = -\frac{\theta^2 - 2N^2}{2N^2} \quad & ; \quad y_1 = -\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2} \\
    x_2 = \frac{\theta^4 - 4N^2\theta^2 + 2N^4}{2N^4} \quad & ; \quad y_2 = \frac{(\theta^3 - 2N^2\theta) \sqrt{4N^2 - \theta^2}}{2N^4} \\
    x_3 = -\frac{\theta^6 - 6N^2\theta^4 + 9N^4\theta^2 - 2N^6}{2N^6} \quad & ; \quad y_3 = -\frac{(\theta^5 - 4N^2\theta^3 + 3N^4\theta) \sqrt{4N^2 - \theta^2}}{2N^6}
\end{align}
$$
By using substitution, it becomes very difficult with $N \geqslant 2$.

The second equation expresses that the points $(x_n,y_n)$ remain on the unit circle, and the first, that the successive points form chords of constant length, subtending an angle $\alpha=2\arcsin\frac\theta{2N}$.
Hence $$(x_n,y_n)=(\cos n\alpha,\sin n\alpha).$$

By the way, the system has in fact $2^n$ distinct solutions, as from every intermediate point, the chord can be drawn in two directions.
