Let $A$ and $B$ points in the plane and $C$ a point in the perpendiclar bisector of $AB$. It is constructed a sequence of points $C_1,C_2,\dots, C_n,\dots$ in the following way: $C_1=C$ and for $n\geq1$, if $C_n$ does not belongs to $AB$, then $C_{n+1}$ is the circumcentre of the triangle $\triangle{ABC_n}$.

Find all the points $C$ such that the sequence $C_1,C_2,\dots$ is defined for all $n$ and turns eventually periodic.

Note: A sequence $C_1,C_2, \dots$ is called eventually periodic if there exist positive integers $k$ and $p$ such that $C_{n+p}=c_n$ for all $n\geq{k}$.
1. **Initial Setup and Definitions:**
   Let \( A \) and \( B \) be points in the plane, and let \( C \) be a point on the perpendicular bisector of \( AB \). We construct a sequence of points \( C_1, C_2, \ldots \) as follows: \( C_1 = C \), and for \( n \geq 1 \), if \( C_n \) does not belong to \( AB \), then \( C_{n+1} \) is the circumcenter of the triangle \( \triangle ABC_n \).

2. **Angle Consideration:**
   Define \( x_n = \frac{m \angle AC_nB}{\pi} \). The sequence \( x_n \) follows a recurrence relation based on the angle \( \angle AC_nB \):
   \[
   x_{n+1} = 
   \begin{cases} 
   2x_n & \text{if } x_n \leq \frac{1}{2} \\
   1 - x_n & \text{if } x_n > \frac{1}{2}
   \end{cases}
   \]

3. **Number Theory Analysis:**
   The problem now reduces to a number theory problem. We need to determine when the sequence \( x_n \) becomes eventually periodic. The key observation is that if \( x_1 \in \mathbb{Q} \), then the sequence \( x_n \) will be eventually periodic.

4. **Denominator Analysis:**
   Consider the denominator of \( x_n \). If \( x_1 = \frac{p}{q} \) with \( \gcd(p, q) = 1 \), we analyze the behavior of the denominator under the recurrence relation. If \( q \) is even, then after a few steps, the denominator will no longer contain any factor of 2, as \( x_n \) is multiplied by 2 in the recurrence relation.

5. **Case When \( q \) is Odd:**
   If \( q \) is odd, the denominator remains the same for all \( x_n \). Denote by \( p_n \) the denominator of \( x_n \). Since \( x_n \leq 1 \) for all \( n \), \( p_n \) can take only finitely many values. Therefore, \( p_n \) is eventually periodic because the recurrence relation depends only on the previous term.

6. **Periodic Behavior:**
   After some steps, the angle \( \angle AC_nB \) can be the same as \( \angle AC_{n+p}B \), but \( C_n \) and \( C_{n+p} \) can be symmetric with respect to \( AB \). It is easy to see that \( C_{n+2p} \equiv C_n \), establishing the periodicity.

The final answer is \( \boxed{ x_1 \in \mathbb{Q} } \).