# Thread: Finding points on a circumference using radius, origin and angle

1. Registered User
Join Date
Mar 2007
Posts
6

## Finding points on a circumference using radius, origin and angle

This is really making me wish I paid attention in Math at school! I'm trying to write a function that will return the x and y coordinates of a point along the circumference of a circle.

I found what I thought was the equation on another forum, but it doesn't seem to be right. Here is my current function, it assumes that the topmost point of the circle is 0º/360º:

Code:
```function GetCircumferencePoints(Radius, OriginX, OriginY, Angle)
{
return [X,Y];
}```
I can tell it's not working because if my origin is 0,0 and my angle is 90º then my Y coordinate should be 0. Is there anyone here who paid attention in Math and could help me?

2. Registered User
Join Date
Mar 2007
Posts
6
Turns out Math.sin and Math.cos return in radians, so the solution was:

Code:
```function GetCircumferencePoints(Radius, OriginX, OriginY, Angle)
{
return [X,Y];
}```
Well, it's here in case I'm not the only coder who's brain occasionally goes out to lunch.

3. Just saw you post, but I'm a bit confused.
it assumes that the topmost point of the circle is 0º/360º:
I think the radian measure starts a the right-most position along a line
x0,y0 ---> x0+r,y0
which would be 0 degrees

The top most part of the circle would be
x0,y0 ---> x0,y0-r (-r assumes you are using screen pixels if plotting)
which is normally considered +90 degrees

I believe x0,y0 ---> x0,y0+r would be the 270 degree position on the circle.

If you really want 0 degrees to be at the top of the circle,
I think you are going to need to add a 'fudge' factor for your calculations.
Should be easy to do by add +90 degrees before you start using the conversion function
and of course subtracting 90 degrees if you need to reverse the process.

Note: This assumes I'm understanding your problem correctly.