## Planar sets closed under intersection of circles

Let $P$ be the plane with a point at infinity. By plane, I mean the Euclidian plane, and therefore it has circles. A line is also a circle, though its center is at infinity. If $A\subset P$ has cardinal $|A|=3$, there exists a unique circle (possibly a line) containing $A$; let me denote it $\Gamma_A$.

Let me say that a subset $X$ of $P$ is circularly stable if it satisfies the following property:

For every subset $A,B\subset X$ with $|A|=|B|=3$ and $\Gamma_A\ne \Gamma_B$, the intersection of $\Gamma_A$ and $\Gamma_B$ is included in $X$.

Every non-colinear $X$ with $|X|\le4$ is circularly stable. A line or a circle is circularly stable, and $P$ itself is so too. If $X$ has a non-void interior, then $X=P$.

Q: What can look like a circularly stable subset $X$ of the plane ? For instance, is it true that if $|X|\ge5$ and $X$ is circularly stable, then $X$ is dense in $P$, or $X$ is a line or a circle ?

Motivation: In classical geometry, one may wander what are the continuous maps $f:P\rightarrow P$ which transform circles or lines into circles or lines. If the guess above is true, then $f$ is completely determined by the images of $5$ non-colinear points.

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Send one of the points to infinity by a Mobius tranformation. Your set of points now has the property that the intersection of any two lines passing from pairs of points in the set is also in the set. Such configurations are either all collinear, dense in the plane, or one of these two exceptional cases:

• A point together with a collection of points in a line
• The four vertices of a parallelogram and the intersection of its diagonals.

The first case cannot be circularly stable as soon as $|X|\geq 6$. This is because if we let $P$ be the apex and $A,B,C,D$ be collinear points in that order, then the circumcircles of $PAC$ and $PBD$ intersect in a point not in $X$. We conclude that if $|X|\geq 7$ and $X$ is circularly stable, then either $X$ is collinear or it is dense in the plane.

To classify all circularly stable sets with $|X|=6$ that are not collinear we must be in the second case above. Let the points be $A,B,C,D$ as vertices of the parallelogram and $O$ the intersection of the diagonals. Now, the circumcircle of $ABC$ and the line $BD$ must intersect at $D$ so $ABCD$ is inscribed in a circle, i.e. is a rectangle. Now looking at the circumcircle of $ABO$ and the line $BC$ we must have $BC$ tangent to this circumcircle so $ABCD$ must be a square. It's easy to check that a square, its center and the point at infinity form a circularly stable set.

There are no circularly stable sets with $|X|=5$ that are not collinear. If there were it would come from the first case. We have the apex $A$ and three collinear points $B,C,D$. The circumcircle of $ABD$ and the line $AC$ intersect in a different point, contradiction. On the other hand, as you mention in the OP, any set with $|X|\le 4$ is circularly stable.

The result I used above is proved in "A dense planar point set from iterated line intersections" by D. Ismailescu and R. Radoicic.

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 @Gjerhji: I think, meant complex-projective (or Moebius) transformation (since real ones do not map circles to circles). – Misha May 15 2012 at 15:22 @Misha, thank you, I changed projective to Mobius :) – Gjergji Zaimi May 15 2012 at 15:27 @Gjergji. Isn't it a problem in the second case ? Let us denote $a,b,c,d$ the points on the rectangle ($ab$, ..., $da$ being edges), then $O$ its center. The intersection of $\Gamma_{abO}$ and $bc=\Gamma_{bc\infty}$ has an other point, not in $A$.Can we really have $|A|=6$ ? Even a $5$-element circularly stable $A$ is not clear to me. – Denis Serre May 16 2012 at 6:15 @Denis, I added a few more details. You're right, a rectangle and it's center is only circularly stable if it is a square. – Gjergji Zaimi May 16 2012 at 17:39 @Gjergji. Im OK with your answer now. The picture with six vertices is interesting. It raises questions. – Denis Serre May 18 2012 at 7:28
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