# How can I find the center point and radius of an arc segment from quadratic equation generated by a curve fitting software

I am trying to understand how to find the radius and center of an arc segment using a curve fitting software that generates an equation using linear regression analysis. It generates equations such as y= a + bx + cx^2 and states the values of a,b and c. The software nicely graphs the best fit line through all the points of the arc and gives statistics as to the accuracy of the fit.

My problem is that I need to be able to calculate the center of the arc and the radius of the line generated by the software. The original data is a table x and y coordinates digitized from known curved surfaces machined from metal using a computerized milling machine. The goal is to digitize unknown curves and solve for the radius and center of those curves. At this point we are using known arcs to understand the math involved.

Unfortunately when I try to randomly select 3 x,y data points using circle formula or even enter the values into an online calculator, the radius values generated are very inconsistent. I am hoping the curve fitting software can do a better job - if I can just understand how to effectively use the output!

Thanks, Keith

• From your description, the software produces a best-fit parabolic arc, not a circular one. Unless it’s very short, you’re going to find a good circular approximation to it.
– amd
Nov 26, 2017 at 1:58

Given that you already have $y=a+bx+cx^2$ as an approximation, then the required center and radius corresponding to the point $(x,y)$ are given by the following procedure:
1. Calculate $p=b+2cx$, $q=2cp/(1+p^2)$, $u=2c-pq$, $v=\sqrt{u^2+q^2}$.
2. The radius is given by $$r=\frac{(1+p^2)^{3/2}}{2c}$$
3. The center has coordinates $$x_0=x-\frac{rq}{v},\quad y_0=y+\frac{ru}{v}$$