Calculating the position of points in a circle?

A point at angle theta on the circle whose centre is (x0,y0) and whose radius is r is (x0 + r cos theta, y0 + r sin theta). Now choose theta values evenly spaced between 0 and 2pi.

Here's a solution using C#: void DrawCirclePoints(int points, double radius, Point center) { double slice = 2 * Math. PI / points; for (int I = 0; I Cos(angle)); int newY = (int)(center. Y + radius * Math.

Sin(angle)); Point p = new Point(newX, newY); Console. WriteLine(p); } } Sample output from DrawCirclePoints(8, 10, new Point(0,0));: {X=10,Y=0} {X=7,Y=7} {X=0,Y=10} {X=-7,Y=7} {X=-10,Y=0} {X=-7,Y=-7} {X=0,Y=-10} {X=7,Y=-7} Good luck!

Given a radius length r and an angle t in radians and a circle's center (h,k), you can calculate the coordinates of a point on the circumference as follows (this is pseudo-code, you'll have to adapt it to your language): float x = r*cos(t) + h; float y = r*sin(t) + k.

The angle between each of your points is going to be 2Pi/x so you can say that for points n= 0 to x-1 the angle from a defined 0 point is 2nPi/x. Assuming your first point is at (r,0) (where r is the distance from the centre point) then the positions relative to the central point will be: rCos(2nPi/x),rSin(2nPi/x).

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