Number of point subsets that can be covered by a disk Given $n$ distinct points in the (real) plane, how many distinct non-empty subsets of these points can be covered by some (closed) disk?
I conjecture that if no three points are collinear and no four points are concyclic then there are $\frac{n}{6}(n^2+5)$ distinct non-empty subsets that can be covered by a disk. (I have the outline of an argument, but it needs more work. See my answer below.)
Is this conjecture correct? Is there a good BOOK proof?
This question is rather simpler than the related unit disk question. The answer to this question provides an upper bound to the unit disk question (for $k=1$).

When $n=6$, consider four points at the corner of a square, and two more points very close together near the center of the square. To be precise, let's take points at $(\pm1,0)$ and $(0,\pm1)$ and at $(\epsilon,\epsilon)$ and $(-2\epsilon,\epsilon)$ for some small $\epsilon>0$. Then if I'm not mistaken, the number of nonempty subsets that can be covered by a disk is 34 (6 of size 1, 11 of size 2, 8 of size 3, 4 of size 4, 4 of size 5, and 1 of size 6), while your conjectured formula gives 41 [I originally had the incorrect 31].
Now that I think about, when $n=4$, taking two points very near the midpoint of the segment joining the other two points (say $(\pm1,0)$ and $(0,\pm\epsilon)$) gives 12 such nonempty subsets (4 of size 1, 5 of size 2, 2 of size 3, and 1 of size 4) while your conjectured formula gives 14.
Edited to add: it's been commented correctly that the above $n=4$ example does indeed give 14, since every size-3 subset can be covered by a disk. But what about if the four points are $(\pm1,0)$, $(0,\epsilon)$, and $(0,0)$ instead? Now I believe the first three points cannot be covered by a disk without covering the fourth also. Perhaps you want to add the condition that no three points are collinear?
