Calculating centre of rotation given point coordinates at different positions I am trying to work out if the centre of rotation of a measured sphere is actually at 0,0 or slightly offset from the centre.
The situation is as follows:
I have a machine tool with a table that rotates about its centre. I wish to verify that the centre of the table is at 0,0 according to the machine coordinate system. I have measured the centre position of a sphere mounted on the table at various rotations of the table so have a number of x,y coordinates for the sphere centre relative to 0,0 (note: the sphere was not necessarily mounted on the table at x=0). Hopefully the image helps to illustrate this.

I would like to be able to calculate the actual centre of rotation (shown by the purple dot) from these coordinates and the angles. 
I have a feeling that there is a simple way of doing this but I cannot work it out.
Any help on this would be greatly appreciated.
Many thanks in advance.
 A: You just need three measurements, say $P_i=(x_i,y_i)$ for $i=1,2,3$.
Let $L_1$ be the perpendicular bisector of $P_1 P_2$ and $L_2$ be the perpendicular bisector of $P_2 P_3$.
The centre is the point where $L_1$ and $L_2$ intersect.
A: This is a problem of circular regression. The answer is in : https://fr.scribd.com/doc/14819165/Regressions-coniques-quadriques-circulaire-spherique
Since this paper is written in French, I joint a copy of the equations to be used :

The radius ($R$) is the radius of the circle fitted to the set of points.
Same method, matrix presentation :

A: For each three points you can find the circle that passes through those points. Mark the center of this circle. Repeat for all the combinations of three points and average the resulting x and y coordinates of the circle.
Look at https://math.stackexchange.com/a/213670/3301 for one of the ways you can get the circle center from three points.
The generelized version of this method is to calculate the following parameters from the coordinates $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$
$$ T = x_1 (y_2-y_3)+x_2 (y_3-y_1) + x_3 (y_1-y_2) $$
  $$ K_2 = (x_1^2-x_2^2)+(y_1^2-y_2^2) $$
  $$ K_3 = (x_1^2-x_3^2)+(y_1^2-y_3^2) $$
Now the circle center is located at
$$ x_c = \frac{K_2 (y_1-y_3)+K_3 (y_2-y_1)}{2 T} $$
$$ y_c = -\frac{K_2 (x_1-x_3)+K_3 (x_2-x_1)}{2 T} $$
With the example provided in the link you get (3,2) for the center

