MHB Finding 2D Polygon Coordinates from a point

[Krotus]

Suppose that I have the coordinates of x and y on a plane.

I am writing a piece of software where the user can select a polygon of 3, 4, 5, 6 or 8 sides. All of the polygon points are equidistant from the x, y point. In other words, if you drew a circle where the center was the x, y point, all of the points of the polygon would line on the circle.

That means, obviously, that the distance of each polygon point is equal to the imaginary circle's radius.

Given that information, what are the equations to create each type of polygon's set of points?

 

[I like Serena]

Suppose that I have the coordinates of x and y on a plane.

I am writing a piece of software where the user can select a polygon of 3, 4, 5, 6 or 8 sides. All of the polygon points are equidistant from the x, y point. In other words, if you drew a circle where the center was the x, y point, all of the points of the polygon would line on the circle.

That means, obviously, that the distance of each polygon point is equal to the imaginary circle's radius.

Given that information, what are the equations to create each type of polygon's set of points?

Hi Krotus, welcome to MHB!

Suppose the polygon will have $n$ sides.
And suppose each of the polygon points must have a distance of $r$ to point $(x,y)$.
Then the x- and y-coordinates of point $k$ of the polygon are given by:
$$\begin{cases}x + r \cos(2\pi \cdot k/n) \\ y + r\sin(2\pi \cdot k/n)\end{cases}$$
where $k$ runs from $0$ to $n-1$. Furthermore, the first point ($k=0$) will be to the right of $(x,y)$.

If you want the first point to be in a different location than to the right of $(x,y)$, we can add a fixed constant to the calls of $\cos$ and $\sin$.

 

[Krotus]

Thanks! Very simple. I knew I had to be overthinking it.