I should have been more precise - the centre I was thinking of is the intersection of the perpendicular bisectors for each side. This finds the point which is the centre of a circumscribing circle. As you say there are other centres, and anyway it is certainly not the correct solution <sigh>.

Coming back to my other idea, take the average of all points. I feels like at least it would be an easy way to make a first guess for an iterative algorithm. In fact without trying it, it feels like the centre of the group would lie on a line passing through the furthest outlying point and the average point.

WRT macros - could be done but don't think this is the most appropriate tool for the job - hard work - embedded VB would be fine.