How to find the radius of a circle given three points in 3d space C#

Question

I have locations of three points along the circle. pt1(x1, y1,z1) , pt2(x2, y2, z2) , pt3(x3, y3,z3) . and want to find the radius of the circle. Already I have a function to compute radius in 2d space, which I am copying it here

public static double ComputeRadius(Location a, Location b, Location c)
        {
            double x1 = a.x;
            double y1 = a.y;
            double x2 = b.x;
            double y2 = b.y;
            double x3 = c.x;
            double y3 = c.y;
            double mr = (double)((y2 - y1) / (x2 - x1));
            double mt = (double)((y3 - y2) / (x3 - x2));

            double xc = (double)((mr * mt * (y3 - y1) + mr * (x2 + x3) - mt * (x1 + x2)) / (2 * (mr - mt)));

            double yc = (double)((-1 / mr) * (xc - (x1 + x2) / 2) + (y1 + y2) / 2);
            double d = (xc - x1) * (xc - x1) + (yc - y1) * (yc - y1);

            return Math.Sqrt(d);
        }

Answer 1

If you know the order of points pt1,pt2,pt3 along the circle then you can use graphical method:

1. cast normal axises from middle of each line segment in the plane of circle

   your circle plane is defined by your 3 points. so the normal vector is

    n = (pt2-pt1) x (pt3-pt2) 

   where the x is cross product so you have 2 lines (pt1,pt2) and (pt2,pt3) in black. The mid points are easy

    p0=0.5*(pt1+pt2) p1=0.5*(pt2+pt3) 

   the axis directions can be obtained also by cross product

    dp0=(pt2-pt1) xn dp1=(pt3-pt2) xn 

   so you got 2 axises:

    pnt0(t)=p0+dp0*t pnt1(u)=p1+dp1*u 

   Where t,u are scalar parameters t,u=(-inf,+inf) it is just position in axis from the starting mid point ...

2. the intersection is center of circle

   So find the intersection of 2 axises and call it pt0

3. compute distance between center and any of your points

    r=|pt1-pt0| 

Sorry the image is for any curve (too lazy to repaint for circle as it is almost the same). If you do not know the order of points then the 2 points that are most distant to each other are the outer points ... In case they are equidistant the order does not matter any is OK