First year french college. Is there a finite set of points of the plane A. Finite set of point A such as... Is there a finite set of at least 3 points in the plane A, such as every 3 points of A are not aligned, and such as every circle going through 3 points of more of A, it’s center is in A ?
I believe that there can’t be such a set. So I tried showing that for every n point of the plane you can’t arrange them to be set A.
Being based on a number n I also tried doing it by doing a « mathematical induction » to show that there can’t be a set A for n points.
I’m a french student in my first year of college, I’m sorry by advance if my english is a bit wacky.
edit: forgot that it's a set of at least 3 points sorry
 A: Answer to be proved in details
Such a finite set of points $A$ can't exist.
Denote by $P_1, P_2, P_3$ three points for which the circumscribed circle $\mathcal C$ is achieving the minimum radius $R$ of all circumscribed circles. Let $O$ be its center. By hypothesis $O \in A$.
Assumption to be proven: one of the triangles $(O \ P_1 \ P_2)$, $(O \ P_2 \ P_3)$ or $(O \ P_1 \ P_3)$ has a circumscribed radius strictly less than $R$. A contradiction.
A: Assume such a set $A$ exists.
Let $X,Y$ be two points in $A$ that are the closest of all pairs of points in $A$.
Look at the bisector $l$ of the segment $XY$. For each point $Z\in A\setminus\{X,Y\}$ we can find the circumcentre of $\triangle XYZ$ and it will be on $l$. Thus, some points in $A$ will be on $l$, and so let's choose the point $O\in l$ so that it is the closest to the line $XY$.

*

*Case 1: $O$ is on the line $XY$, and so $A$ contains three colinear points - a contradiction.

*Case 2: $O$ is not on the line $XY$. Note that $OX=OY\ge XY$ so the angle $\angle XOY$ (opposite to the shortest side $XY$) is acute, and so all three angles of the isosceles triangle $\triangle OXY$ are acute. This means that the circumcentre $O'$ of $\triangle XOY$ lies inside $\triangle XOY$, and therefore $O'\in l$ and is closer to $XY$ than $O$ - a contradiction.

A: Assuming $\text{Card} A\ge 3$, let $P_1, P_2, P_3$ be three points in $A$ such that the circle passing through the three points has maximal radius. Let C be the center of this circle. If any of the angles $P_i C P_j$ is $> 2\pi/3$, then the circle passing through $P_i, C, P_j$ has a larger radius, a contradiction. It follows that all the angles $P_i C Pj$ are equal to $2\pi/3$, but then the center of the circle passing through $P_1, C, P_2$ is aligned with $C$ and $P_3$, another contradiction. Hence there can be no such set with $\text{Card} A \ge 3$.
A: In the plane draw a circle, take  three points on the circle (I think you can take any  finite number of points but 3 will do). Call these points your set A. These 3 points are not aligned (not on the same line) and there is only one circle that goes through them and we know by construction that this circle is in the plane.
