Putting points into groups of three to find their centers

For example, if I have points A, B, C, D, how would I be able to find the centers of circle ABC, ACD, ABD, and BCD?

I found a function to get the centers from circles: https://www.mathworks.com/matlabcentral/fileexchange/57668-fit-circle-through-3-points

But I do not know how to find all the combinations of three points from my data set. I thought the nchoosek function might work, but I have my data set includes both x and y values for each point and they are not intergers. Would appricate any suggestions!

[Jan]

Please post some code which defines the inputs. "points A,B,C,D" does not tell us, in which format you save the coordinates: [1 x 2], [2 x 1], fields "x" and "y" of a struct, etc.

[Jonathan Soucy]

Hey Jan, here is a sampling of the inputs I am working with

XYgroups = [3897.8, 835.0;

3896.6, 875.1;

2562.4, 979.4;

2526.8, 975.1

2500.1, 944.2]

These arrays can be >100 in length

Answer 1

[Jan]

Guessing, that the "points" are [1x2] vectors:

A = [1, 12];
B = [2, 10];
C = [3, 17];
D = [4, 9];

P = cat(1, A, B, C, D);  % Much better than a set of variables!
% Now you can usen an index:
Groups = nchoosek(1:4, 3);

Pairs = 4×3

1 2 3 1 2 4 1 3 4 2 3 4

for k = 1:size(Gropus, 1)
    aGroup = Gropus(k, :);
    ThreeP = P(aGroup, :);  % Coordinates of 3 points as [3 x 2] matrix
    ... Do what you want with them
end

Idea: Do not create a set of variables, but store them in a matrix directly. Then it is trivial to access them using an index.

[Jonathan Soucy]

thank you! Worked perfect! Here's my final code

XYvals = [cell_VXYZ(:,2),cell_VXYZ(:,3)];
groupindex = nchoosek(1:length(XYvals),3);
Centers = [];

for k = 1:size(groupindex, 1)
    tempgroup = groupindex(k, :);
    tempXYvals = XYvals(tempgroup, :);
    [~,tempCenters] = fitcircle3(tempXYvals);
    Centers = [Centers; tempCenters']; 
end

[Image Analyst]

Why do just a subset of the numbers? Why not do them all? A hundred or more is not very large. Even a million wouldn't be. If you have tens of millions though, it might take a bit of time.