If you have a polygon with equal sides and equal distance from center to all vertices it seems to be a regular convex polygon
EDIT I've found much easier way.
Assume that center of the polygon has coordinates (x_0,y_0) and known vertice has coordinates (x_n,y_n). Also assume that we are considering n-sided polygon.
Coordinates of i-th vertce (0<i<n) can be calculated using this formulae
x_i = x_0+R*cos(a+2*Pi*i/n)
y_i = y_0+R*sin(a+2*Pi*i/n)
where
_______________________
R = v(x_n-x_0)^2+(y_n-y_0)^2
a = acos((x_n-x_0)/R)
According to your example computations using formula above shows that
A=(4, 6)
B=(0.1339745962155614, 6.2320508075688776)
C=(1.8660254037844377, 2.7679491924311228)
You can check (e.g. using this calculator) that distances between A and B, B and C, C and A are the same and equal to 3.8729833462074166. Also yu can calculate distance between center and each vertice and see that they all will be the same.
==================
That means you can find length of the side of such polygon using this formula
a=2Rsin(Pi/n), where R is a distance between center c of your poly and its known vertice p.
_____________________________
R=v(c.x - p.x)^2 + (c.y - p.y)^2
So you will have a triangle based on center c of your poly and its first p and second s vertices. Since you know coordinates of c and p and length of all sides of this triangle (distance between c and p is R, distance between p and s is a and distance between s and c is again R) you can determine coordinates of s.
For simplicity I am assuming the centre of the polygon is origin.
If two consecutive vertices are given then the angle they make at origin can be found. Call it $\theta=2\pi/n$. Now all points are obtainable from any single point by rotating repeatedly by this angle (centred at origin). As rotation is a linear transformation, this can be achieved by matrix multiplication
$$ R_\theta=\pmatrix{\cos\theta & -\sin\theta\cr \sin\theta &\cos\theta\cr}$$
Write the co-ordinates of a vertex of the polygon as a column vector $v_1=\displaystyle {x_1\choose y_1}$. Matrix multiplication $v_2=R_\theta v_1, v_3=R_\theta v_2$ etc will give the co-ordinates of all the vertices.
Best Answer
Find the equation of the line for two of the medians, and compute the intersection.
The line from $v_1$ to the midpoint of the opposite side can be parametrized by:
$$(x,y,z)=(0,0,0)+t(2,1,3)$$
The line from $v_2$ to the midpoint of the opposite side can be parametrized by:
$$(x,y,z)=(0,1,1)+s(1,-1,0)$$
This gives us the system of equations:
$$x=2t=s\\y=t=1-s\\z=3t=1$$
So we see that $t=\frac{1}{3}$ and $s=\frac{2}{3}$ is the intersection, which gives the point $(\frac{2}{3},\frac{1}{3},1)$ as the centroid.