You can use my interdists() function below. Since you say it is a circle, I don't see why you would want to compare all pairs of points. The only way this can give you a good approximation of the diameter is if the circle is well-sampled, and in that case, the distance to one particular point ,e.g.,
data=[x;y];
distances = interdists(data(:,1),data);
ought to be enough.
function Graph=interdists(A,B)
%Finds the matrix of distances between point coordinates
%
% (1) Graph=interdists(A,B)
%
% in:
%
% A: matrix whose columns are coordinates of points, for example
% [[x1;y1;z1], [x2;y2;z2] ,..., [xM;yM;zM]]
% but the columns may be points in a space of any dimension, not just 3D.
%
% B: A second matrix whose columns are coordinates of points in the same
% Euclidean space. Default B=A.
%
%
% out:
%
% Graph: The MxN matrix of separation distances in l2 norm between the coordinates.
% Namely, Graph(i,j) will be the distance between A(:,i) and B(:,j).
%
%
% (2) interdists(A,'noself') is the same as interdists(A), except the output
% diagonals will be NaN instead of zero. Hence, for example, operations
% like min(interdists(A,'noself')) will ignore self-distances.
%
% See also getgraph
noself=false;
if nargin<2
B=A;
elseif ischar(B)&&strcmpi(B,'noself')
noself=true;
B=A;
end
N=size(A,1);
B=reshape(B,N,1,[]);
Graph=l2norm(bsxfun(@minus, A, B),1);
Graph=squeeze(Graph);
if noself
n=length(Graph);
Graph(linspace(1,n^2,n))=nan;
