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This is problem 12093 of the American Math Monthly, published a few months back:

Let $S$ be a finite set of points in the plane no three of which are collinear and no four of which are concyclic. A coloring of the points of S with colors red and blue is circle- separable if there is a circle whose interior contains all the red points of $S$ and whose exterior contains all the blue points of $S$. Determine the number of circle separable colorings of $S$.

It seems the number of colorings is, remarkably, independent of the configuration of the points. Let $n=|S|$. In addition to the trivial colorings where red circles contain only one point or none at all (in total $(n+1)$ of them), it seems that every pair of points uniquely determine a circle-separable coloring, and every three out of these $n$ also determine uniquely a circle-separable coloring.

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  • $\begingroup$ 1. Do you have a question? 2. I have one: what do you mean by the last two clauses of the last sentence? $\endgroup$ – Peter Taylor Jul 4 at 14:21
  • $\begingroup$ The question is stated above; while I have an answer to it, the question seems interesting enough to share, in order to possibly see different approaches. I edited the last sentence to hopefully make it more clear. $\endgroup$ – Aritro Pathak Jul 4 at 14:29
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    $\begingroup$ What circle contains $(0, -100)$ and $(0, 100)$ but neither of $(-1, 0)$ or $(1, 0)$? $\endgroup$ – Peter Taylor Jul 4 at 15:03
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    $\begingroup$ I'm having second thoughts about that example, actually. The circle which has those two points as a diameter contains both of the other points, and so isn't counted under any other heading. $\endgroup$ – Peter Taylor Jul 4 at 16:23
  • $\begingroup$ As you allude to above; you do have ${{4} \choose {0}}+{{4}\choose {1}}+{{4}\choose{2}}+{{4}\choose{3}}$ colorings; and exactly one out of the ${{4}\choose{2}}$ colorings is one where al four points are colored red. $\endgroup$ – Aritro Pathak Jul 4 at 17:39
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Let the number of circle seperable colourings for $n$ points be $f_n$. By rearranging, we see that to show $f_n$ is as described, we just need to show $f_{n}-f_{n-1}={{n}\choose 2}+1$

For this, fix a point $P$, and note that we have a bijection between $n-1$ circle seperable colourings without $P$, and $n$ circle seperable colourings where $P$ has no choice in its colour. Thus, we may identify $f_{n+1}-f_n$ with the number of CSCs where $P$ has a choice in its colour. The space of circles/lines which seperate two sets of points in the plane is connected, so we have that if $P$ has a choice in its colour given a CSC on the rest, it can be taken to lie on the seperating circle (intermediate value theorem), and perturbing the circle yields the two choices of colour.

Given a partition of the points without $P$ admitting a seperating circle passing through $P$, then sometimes we can colour the partition red/blue in both ways, if our circle's interior can be taken to contain either part of our partition. This is determined by the side on which $\infty$ is on, viewing this as taking place on $\mathbb{CP}^1\cong S^2$. So apply a mobius transformation $\mu$ taking $P$ to $\infty$, now we have $n-1$ points in the plane, and a new point $\mu(\infty)$, and CSCs admitting a circle passing through $P$ are just lines seperating this set. This set has no three points collinear by our hypothesis, and thus it suffices to show that the number of seperating lines in ${n\choose 2}+1$. But this is a simple induction, adding an extra point $Q$ going from $n-1$ to $n$ adds $n$ new seperating lines, corresponding to the $n$ potential seperating lines that pass through $Q$.

There is also probably a cleverer non inductive proof for this last point, but I think the main trick is to mobius transform a point to $\infty$.

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