TITLE: A problem with concyclic points on $\mathbb{R}^2$
QUESTION [21 upvotes]: 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

REPLY [5 votes]: 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.