I am trying to prove the following statement. Any suggestions or references are highly appreciated.
Consider $n$ points in $R^2$, i.e., $x_i\in R^2, i=1,\ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $\bar{x}=\frac{1}{n}\sum_{k=1}^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,\ldots,n$), then there exists $\alpha\in R$ such that \begin{align} x_{i+1}-x_i + x_{i-1} - x_i & = \alpha (\bar{x}-x_i), \forall i=2,\ldots,n-1, \\ x_{2}-x_1 + x_{n} - x_1 & = \alpha (\bar{x}-x_1), \\ x_{1}-x_n + x_{n-1} - x_n & = \alpha (\bar{x}-x_n). \end{align}
This does not look true as stated
Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $\bar{x}=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:
$(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) \qquad$ $= \frac25\big((0,0)-(-5,0)\big)$
$(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) \qquad$ $ \not= \frac25\big((0,0)-(-4,-3)\big)$