Convert a circle to a polygon If I have a circle with a given radius, how calculate I find a regular polygon's vertices, if I know they are all on the circles edge (like in the images below)?

 A: These points on the unit circle given by $x^2+y^2=1$ will form a regular $n$-gon:
$$(x,y)=(\cos(\frac{2k\pi}{n}),\sin(\frac{2k\pi}{n}))$$
for $k=0, 1,\cdots, n-1$.
And they're in fact the roots of unity which means the complex roots of the equation $z^n=1$. You need to know complex numbers to fully understand this answer.
But you can also use trigonometry to achieve this. Note that you want to divide a circle into $n$ sections each of $\frac{360}{n}$ degrees. Now use trigonometry to see why that's true. When $k$ ranges from $0$ to $n-1$, you are moving $\frac{360}{n}$ degrees forward on the circle. Does it make sense?
A: For a polygon with $n$ vertices (an $n$-gon), each vertex will have an angle of $\frac{2\pi}{n}$ between it and the next one. All we need to do is place a first vertex, then translate it around the origin by this amount $n$ times to get all the vertices.
Place the first vertex at $p_0=(\rho, 0)^T$, where $\rho$ is the circle's radius. To rotate a point by $\theta$ around the origin we need the rotation matrix in 2 dimensions, $R(\theta)$ given by:
$$
R(\theta)= \left( \begin{matrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{matrix} \right)
$$
Just apply this multiple times to get each vertex:
$$
p_m=R^m \left( \frac{2\pi}{n} \right) p_0
$$
So the set of vertices for a certain $n$-gon is given by:
$$
\{R^m\left(\frac{2\pi}{n}\right)p_0 | p_0=(\rho, 0)^T, 0\le m \lt n\}
$$
A: If you are not looking for exactness as an algebraists and merely looking for high precision engineering solution, it is just a matter of looking up the trigonometric table once.
For example, to get the vertices of 11-sided polygon: we need to find 11 equally distant points on the circle of given radius $r$: call the points $P_1,P_2,\ldots, P_{11}$.
$P_1$ will be the point on the positive $x$-axis lying on the circle i.e., $(r,0)$. Then $P_2$ will be the point with co-ordinates $(r\cos \frac{360^\circ}{11}, r\sin \frac{360^\circ}{11})$. Now you know the distance between $P_1$ and $P_2$ use this to mark $P_3$ (that is $P_3$ would be on the circle and at the same distance as $P_1$ is from $P_2$) and so on,
