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 $k=1$).

Here’s a brief outline of the approach I originally took.
The $n\choose 2$ lines that are equidistant from a pair of points partition the plane into 
$$r_n\;=\;{\!\!{n\choose 2}+1\choose 2}+1-{n\choose 3}$$
regions.
For example, five points gives 46 regions:
$\hspace{1.5in}$ 
Each region $\mathcal{R}$ defines a permutation $P_\mathcal{R}$ of the points, ordered by their distance from any point in $\mathcal{R}$. Disks centred on a point in $\mathcal{R}$ cover (the points in) some prefix of $P_\mathcal{R}$.
If $\mathcal{R}$ and $\mathcal{S}$ are two adjacent regions, then $P_\mathcal{R}$ and $P_\mathcal{S}$ differ by the reversal of a single pair of adjacent points (e.g. $\ a\dots pq\dots z\ $ and $\ a\dots qp\dots z\ $). Disks centred on points in $\mathcal{S}$ thus only cover a single subset ($\;a\dots q\;$) of points that is not covered by disks centred on points in $\mathcal{R}$.
This gives us an upper bound of $n+r_n-1$
on the number of coverable subsets: an initial region gives us $n$ subsets; the other $r_n-1$ regions each add at most one further subset.
However, if we consider the permutations corresponding to the regions (in a ‘circuit’) around a point at which (two or three) lines intersect, we see that there is actually one fewer distinct prefix (= coverable subset) for each of the 
$$i_n\;=\;{\!{n\choose 2}\choose 2}-2{n\choose 3}$$
intersection points. [It is this claim that remains to be proved rigorously.]
Thus we have a total of $n+r_n-1-i_n$ (which is $\frac{n}{6} (n^2 +5)$) coverable subsets.
