LabVIEW
Smallest circle around N points
Solved!
A few pictures may help-- and some clarification
In this frame we see how I chose the arbitrary median of [C] to determine the inner boundry of Orbit point O and its arbitrary starting point. Simply O only needs to be outside the light green circle and point out there will do.
Here I have constructed the "chords of interest" and identified all points on the circle (and some "trivial" points) (in this ex 1 point is trivial (that is inside the circle)
Nex we examine the angles of the chords at the vertices. They will either be obtuse, right or accute If the angle is obtuse (as @2) a circle drawn through them will be too small as seen here (because we have found the smallest circle containing those three points 
However If the angle is RIGHT Or accute as seen @1 we can draw the circle though these three points at by selecting the correct origin shown below
The trick to this is that the smallest circle solution must not have all of its points on the circumferance in one hemisphere (except at the special case where excatly two pointsare on the circumference ) . Rotate the top point in [C] CCW on the circumference. the "Chord of interest" rotates rotates CCW at half the radiial velocity and angle 1 becomes more accute BUT, the intesection of the bisecting lines does not change (although at some point a Flip occurs as the point is uneclipsed earlirer in the Orbit)
Perhaps the way to get to this the easiest:
Triangles
- ABC
- ABD
- ACD
- BCD
Are formed from the points we found.
- the circle defined by Obtuse triangle ABC will not enclude point D Refer to your Trig notebook: an obtuse triangle can be inscribed inside a single hemishere of a circle a right triagle has exactly pi radians) and we noted earlier that the arc between all points on the circumferance must be >=pi*r
- Likewise Obtuse triangle BCD will not enclude point A
However, Acute triangles ABD and ACD do meet the the Arc >= pi*r requirement and circles inscribed about Them will define your smallest circle
Darin,
The weakness of my approach lies in determining the trivial Points. in this example, Points C and D are "trivial" in that they form the obtuse angles ACB and ADB. And thusly they do not define the circumference. Although, it would be easy to construct an example where a "Trivial" point IS on the circumferance, the circle would be described sufficiently by the remaining points hence if ANY point forms any obtuse angle it can, and should, be dismissed from further analisys.
Your
implementation DOES solve for this special case. My gut analisys last
evening suffered from lack of proper investigation. Your methodology of
bisecting a triangle in "Circle from diameter.vi" is Briliant! (but
really needs some documentation 
I now suspect your occasional "hangs" are related to the approach your use in step three at Acute Triangle.vi. While I tested a few cases where the origin lies in line with two points and did not observe any "hanging", I suspect the dissmissal of right trianlges may be your culprit. Rightness is Very difficult to determine in a cartisean system limited to DBLs. If you have any examples you've seen hang please post them. Another possibile culprit is in Step two. I suspect there could be cases where two identical points (or two points with identical "Max Distances" found from Check Points.vi) could infinetely swap back and forth between index 0-2.
The only (ERROR?) I have found is the special case where Index[C]=1 :r=0 is correct in that a circle of "no" size contains a point however, Xc =0and Yc=0 places the null circle in the wrong location. Xc=X[0] and Yc=Y[0] in this case.
