Coordinates of a point on regular polygon perimeter Having an angle $\alpha$ between the $y$ axis and a line intersecting the origin, natural number $n$ being the number of sides of regular polygon, radius $R$ and assuming the bottom side of the polygon is parallel to the $x$ axis, how can I calculate coordinates of the point $I$?
Angle $\alpha$ being $15°$ in the picture is just an example. 
 
 A: With this from wikipedia

The circumradius R from the center of a regular polygon to one of the
  vertices is related to the side length s or to the apothem a by
$$     { R={\frac {s}{2\sin \left({\frac {\pi
 }{n}}\right)}}={\frac {a}{\cos \left({\frac {\pi }{n}}\right)}}}
 $$

you can find the length $a$ from the origin to the side of your polygon. That gives you the $y$ coordinate of $I$. For the $x$ coordinate use $\tan \alpha$.
Note that if $n$ is odd the "bottom edge" of the polygon won't be horizontal if one vertex lies on the $x$ axis.
A: Since a regular hexagon has internal angles of $120^\circ$, the sides, except those parallel to the $x$-axis, will be defined as follows (with $r$ being the radius of the circle):
$$y_{1,2,3,4}=\tan\left(\frac{\pm2\pi}{3}\right)(x\pm r)\tag{1}$$
Consider the lone side  $y=\tan\left(\frac{2\pi}{3}\right)(x+r)$ below:

The line intersects the circle at $\left(-\frac{r}{2},-\frac{1}{2} \left(\sqrt{3} r\right)\right)$, which tells us that from the original diagram, $I$ lies on the line: $$y=-\frac{1}{2} \left(\sqrt{3} r\right)\tag{2}$$

The line through the origin is then defined as:
$$y=(\cot\alpha )x\tag{3}$$
where $\alpha$ is your angle. Just then solve for the intersection of $(2)$ and $(3)$.

In your case, for $\alpha=-\frac{\pi}{12},$ the intersection should be:
$$\bbox[10px, border:1px solid red]{\therefore I=\left(\frac{\sqrt{3}r}{2\left(\sqrt{3}+2\right)},-\frac{1}{2}\left(\sqrt{3}r\right)\right)}$$
