A problem with concyclic points on $\mathbb{R}^2$ I am thinking about the following problem: 
If a collection $\{P_1,P_2,\ldots,P_n\}$ of $n$ points are given on the $\mathbb{R^2}$ plane, has the property that for every $3$ points $P_i,P_j,P_k$ in the collection there is a fourth point $P_l$ in the collection such that $P_l$ is con-cyclic with $P_i,P_j,P_k$, (i.e. $P_l$ lies on the circle passing through the points $P_i,P_j,P_k$), does it follow that all the points are necessarily con-cyclic ? 
I would really appreciate if someone finds a proof with basic Euclidean Geometry.
I would call a class of Convex Geometric figure (upto Homothety) on $\mathbb{R}^2$, $k$-determined if exactly $k$ points are required to determine the figure uniquely. For example a circle is $3$-determined, one needs exactly $3$ points on the plane to determine a circle uniquely. An ellipse is $4$-determined.
From here I would like to ask the following question : If a collection $S$ of $n$ points on $\mathbb{R}^2$, has the property that every sub-collection $T_i=\{P_{i_1},\ldots,P_{i_k}\}$ of $k$ points of $S$ has the property that there is a $k+1^{th}$ point, $P_i \in S\setminus T_i$ (distinct from the sub-collection $T_i$) that lies on the $k$-determined convex figure, determined by $T_i$, then  does it follow that all points of $S$ lie on the $k$-determined convex figure?
Inspired from The Sylvester-Gallai Theorem
 A: The conjecture is true, and can be recovered from the standard
Sylverster-Gallai theorem in the plane.
We first prove:
Lemma 1 Suppose $C$ is a finite configuration of points
in real 3-space ${\bf R}^3$ such that for every $Q_1,Q_2,Q_3 \in C$
there is a plane meeting $C$ in $Q_1,Q_2,Q_3$, and at least one more
point of $C$.  Then $C$ is contained in a plane.
(Of course the plane is unique unless $Q_1,Q_2,Q_3$ are collinear.)
Proof of Lemma 1: fix $Q_1 \in C$ and let $\Pi_1$ be a plane
containing $Q_1$ but no other point of $C$.  Let $\Pi \neq \Pi_1$
be any plane parallel to $\Pi$, and let $C'$ be the set consisting
of the projections from $Q_1$ to $\Pi$ of all $Q \in C$ other than $Q_1$
(that is, the intersections with $\Pi$ of the lines $\overline{Q_1 Q}$).
Applying the hypothesis only to triples that contain $Q_1$
shows that $C'$ satisfies the hypothesis of the Sylvester-Gallai theorem.
Hence $C'$ is contained in a line, whence $C$ is contained in the plane
spanned by this line and $Q_1$.  $\ \diamondsuit$
We connect this with the problem at hand using the following observation:
Lemma 2 Points $(x_i,y_i)$ in the plane are concyclic or collinear
iff the corresponding points $(x_i, y_i, x_i^2 + y_i^2)$ on the
round paraboloid $z = x^2 + y^2$ are coplanar.
Proof of Lemma 2: the intersection of $z = x^2 + y^2$ with any plane
$A_0 + A_1 x + A_2 y + A_3 z = 0$ projects to the locus of
$A_0 + A_1 x + A_2 y + A_3 (x^2+y^2) = 0$, which is line if $A_3 = 0$
and a circle otherwise.  $\ \diamondsuit$
Now assume that $S \subset {\bf R}^2$ is a finite set of points
such that for every $P_1,P_2,P_3 \in S$ there is a circle meeting $S$ in
$P_1,P_2,P_3$, and at least one more point of $S$.  Then by Lemma 2
the associated configuration
$$
C_S := \{ (x,y,x^2+y^2) \in {\bf R}^3 \mid (x,y) \in S \}
$$
satisfies the hypotheses of Lemma 1.  Hence $C_S$ is contained in a plane.
Applying Lemma 2 in reverse, we conclude that $S$ is contained in a circle
or line.  The line does not satisfy the hypothesis, so $S$ is concyclic,
as desired.  QED
The same linearization trick applies to other such problems.
Lemma 1 generalizes to ${\bf R}^d$ for any $d>2$: a finite configuration $C$
such that every $d$ points are on a hyperplane through a $(d+1)$st point
must be contained in a hyperplane; this is proved by projection to a plane
from any $d-2$ points of $C$ in general linear position.  So, for instance,
if $S \subset {\bf R}^2$ has the property that every five points are
on a conic that contains a sixth point of $S$ then the associated
configuration
$$
\{ (x,y,x^2,xy,y^2) \in {\bf R}^5 \mid (x,y) \in S \}
$$
(which lies on the affine version of the
Veronese surface)
is contained in a hyperplane, whence $S$ is contained in a conic.
A: This follows directly by applying Sylvester-Gallai Theorem and inversion.
Consider any collection of $n+1$ points $\{ P_1, P_2, \ldots P_{n+1}\}$. Fix $P_{n+1}$, and apply inversion (with respect to a unit circle) to the remaining $n$ points to obtain $\{Q_1, Q_2, \ldots Q_{n}\}$. Note that $Q_{n+1}$ is the point at infinity, which lies on every line.
Then, Sylvester-Gallai tells us that for the points $\{Q_1, Q_2, \ldots Q_{n}\}$, either
1) all points are collinear or
2) there is a line with exactly 2 points.
Applying the inversion again, we get that
1) all points are concyclic or
2) there is a circle with exactly 3 points.   
Hence, if there is no circle which contains exactly 3 points, then all points are concyclic.
A: This is a very partial answer, just for a few cases ($n=4,5,6,7,9$). Consider all the possible circles that arise in the problem, and let $k$ be the largest number of points from the set that are on one circle. In other words, no circle has more than $k$ points from the set.


*

*Case $k=4$. Say $P_j$ for $j=1,2,3,4$ are on one of the maximal circles (actually, in this case, all circles will have exactly $4$ points), call it $\mathcal{C}$. Let $P_5$ be outside this circle. Then $P_5P_1P_2P_l$ must be on one circle. Moreover, $l$ has to be different than $3$ and $4$, as otherwise $P_5$ would be on $\mathcal{C}$ as well, contradicting the maximality of $k=4$. This solves the problem for $n=5$.
If we have an additional point $P_6$, the previous considerations show $P_5P_1P_2P_6$ must be on the same circle. Similarly, $P_5P_1P_3P_6$ must be on the same circle. That is again impossible, since the previous two circles have $3$ points in common (so they are actually the same) and $5$ points on the circle, again contradicting the maximality of $k=4$. This solves the problem for $n=6$. (Well, almost so far, but we'll see later that the cases $k> 4$ easily work, or rather don't work, in this case as well.)
Similar argument works for $n=7$. In a sense, there are two few points outside, and this is the argument for $k>4$. Essentially you must have at least $k$ points not on the $\mathcal{C}$, in order not to contradict the maximality of $k$.   For $n=8$ the argument no longer works.
For $n=9$, and in fact for any odd number, there is a slightly different argument. If $P_a$, $P_b$, and $P_c$ are three points not on the maximal circle $\mathcal {C}$, we cannot have $P_a P_1 P_2 P_b$ and $P_aP_1P_2P_c$ at the same time (contradicts maximality of $k=4$). Therefore, the  circle containing $P_1$, $P_2$ and a point not on $\mathcal{C}$, must contain exactly one other point not on $\mathcal{C}$. But this is not possible if there are an odd number of points not on $\mathcal{C}$, since they "match" with $P_1$ and $P_2$ on pairs of two. This does not solve the problem for any odd number of points, since we still are in the restrictive case $k=4$. However, the case $n=9$ will not work for $k>4$ as well. 

*Cases $k>4$. By an argument similar to the first part of the case $k=4$, we need at least $5$ points not on $\mathcal{C}$ in order not to contradict the maximality of $k=5$. This solves, as claimed, the cases $n=6$, $n=7$, and $n=9$. However, not $n=8$ since this remained unsolved in the case $k=4$.
That's about it. Note this argument is entirely combinatorial. I couldn't crack $n=8$, all I can say is that IF this case is possible in a non-trivial way (that is, all the points are indeed con-cyclic) it must happen in the case $k=4$. I also tried to build such an example , but I could not. Perhaps there is a geometric argument, rather than combinatorial, why this is not possible. Hope this makes sense.     
A: Don't know if a proof with induction meets your requested answer... But by induction this (seems) to be pretty easy to show:
Proof:
Let $P = \{P_1,\ldots,P_n\}$ be a set of $n$ distinct points in $\mathbb R^2$ with the precondition, that $\forall a,b,c \in \{1,\ldots,n\} \exists d \in \{1,\ldots,n\}$ such that $P_a,P_b,P_c,P_d$ are con-cyclic.
(Induction start): If $n = 4$ then $P_1,P_2,P_3,P_4$ and hence all points are con-cyclic.
(Induction step): Now suppose that for all sets of size $n$ with the desired property the whole set is con-cyclic and let $P$ be such a set with $n$ Points $P_1,\ldots,P_n$. Denote the circle that contains the points from $P$ by $C_1$. Now add another point $P_{n+1}$ to the set. There are two possible cases:


*

*The point $P_{n+1}$ lies on the circle $C_1$. In this case we are done.

*The point $P_{n+1}$ does not lie on the circle $C_1$. Denote the circle containing $P_1,P_2,P_{n+1}$ by $C_2$. Then $P_1,P_2 \in C_1$ and $P_1,P_2 \in C_2$ and hence $C_1 \cap C_2 \supset \{P_1,P_2\}$. But two distinct circles in $\mathbb R^2$ cannot have more than two common points and therefore $C_1 \cap C_2 = \{P_1,P_2\}$. So one cannot find a fourth point $P' \in P$ such that $P_1,P_2,P_{n+1},P'$ are con-cyclic. This is a contradiction and hence the point $P_{n+1}$ has to lie on $C_1$.


Open problem: Does every set of $n+1$ points with the desired property arise from a subset of $n$ points with the same property plus a new $n+1$th point?
$k$-determined convex figures
The more I think about the special case with the circles, the more I believe, that not the $k$-deterministed property is relevant, but how many shared points two distinct figures can have at most.
Two distinct circles can have at most two common points, hence having for each 3-tuple of points on a circle a fourth point, that is on the circle, too, makes all points lie on one circle.
Two distinct lines can have at most one common point, hence having for each pair of points on a line a third point, thats is on the same line, too, makes all points lie on one line.
For ellipses your assumption would need a fifth point for every 4-tuple of points (since ellipses are 4-determined). But I guess, that since two distinct ellipses can share at most 4 common points, it would need a sixst point for every 5-tuple of points on the same ellipse.
Edited:


*

*Rewrote induction step 2 to be more clear.

*Add idea for the more general question.

