Real World CNC Machining Geometry Problem I need to re-find center on a circle. In our tool room a tool maker milled an annulus with a 4 hole pattern. The holes were milled in a square, each equidistant from the center of the annulus. The part was removed from the machine and pins were installed into the holes. Then returned to the machine for more work.

In order to re-find the center the tool maker puts a dial indicator in the spindle and moves the spindle close to the center of the part and checks the pins with the indicator. The indicator will tell him the relative distance each pin is from the center of the spindle. in other words the indicator tip is some at some arbitrary radius from the center of the spindle and when it moves past a pin the dial shows how much closer the pin is to center than the neutral position of the tip.
Known:

*

*The radius of the annulus, pin pattern and pins

*The origin of the coordinate system is within the annulus

*The origin of the coordinate system is defined by the axis of the spindle/indicator.

*The difference, radially, between a radius of constant, but arbitrary length and the distance from the pins and the origin of the coordinate system.

Unknown

*

*The angle of the square hole pattern to the coordinate system

*Location of the center of the annulus relative to the coordinate system.

Assumptions:

*

*The parts are machined to perfect tolerances

Looking for:

*

*An equation that describes the center of the annulus

Update:
Please forgive the hand drawn diagram (if anyone has a recommendation for an easy way to draw pretty diagrams on the computer, please let me know).
Here is a drawing of how I visualize the problem.
R is the arbitrary, unknown radius of the indicator centered at the origin.
$\delta_A$ (etc.) is the measurement from the indicator. Therefor, point A (the closest point of the pin to the center) is at a distance of R-$\delta_A$ from the center, with an unknown angle $\theta_A$
Square ABCD is of known size but not known position, so I cannot simply find it's center. AB=BC=CD=AD, AC=BD; known: length AC
Looking for the position of center point c (sorry about any ambiguity around pin C and center point c)

 A: 
Let (known) common length $CA_k=\rho$.
Let $r$ be the length and $\theta$ the polar angle of $\vec{OC}$.
Let $\rho$ be the length and $\alpha$ the polar angle of $\vec{CA_4}$.
With the other notations of the figure, we can write 4 equations with 3 unknowns, each one being  the cosine law applied to one of the four triangles with "a" red, "a" blue and "the" green arrow :
$$OA_k^2=OC^2+CA_k^2-2 OC.CA_k \cos(...)$$
i.e.,
$$\color{blue}{\gamma_k}^2=\color{green}{r}^2+\color{red}{\rho}^2-2 r \rho \cos(\theta+\alpha+k \pi/2), \ \ \ \ k=1,2,3,4,$$
giving rise to a system of four equations with three unknowns $r,\theta$ and $R$:
$$\begin{cases}
k=1: \ \ &(R-\delta_1)^2=\gamma_1^2&=&r^2+\rho^2+2 r \rho \sin(\theta)\\
k=2: \ \ &(R-\delta_2)^2=\gamma_2^2&=&r^2+\rho^2+2 r \rho \cos(\theta)\\
k=3: \ \ &(R-\delta_3)^2=\gamma_3^2&=&r^2+\rho^2-2 r \rho \sin(\theta)\\
k=3: \ \ &(R-\delta_4)^2=\gamma_4^2&=&r^2+\rho^2-2 r \rho \cos(\theta)
\end{cases}$$
which looks tractable ; for example
$$\frac{Eq. 1-Eq. 3}{Eq. 2-Eq. 4} \iff \tan \theta = \frac{\gamma_1^2-\gamma_3^2}{\gamma_2^2-\gamma_4^2}$$
connecting in fact angle $\theta$ and radius $R$.
Anyway, there is a good hope that this system having less unknowns than equations is solvable...
I stop there and ask you if it is the kind of development you wanted or not, or not exactly...
