Least-Squares Circle Fit Given a finite set of points in R2, say { (x i,yi ...

October 24, 2006 10:22 am MDT Page 2 of 3
Expanding Eq. 2 gives
∑ [
u i u
2
i − 2 u i u c + u 2 c + vi 2 − 2 v i v c + vc 2 − α ] = 0
i
Defining S u = ∑ i u i, S uu = ∑ i u2 i , etc., we can rewrite this as
S uuu − 2 u c S uu + u 2 c S u + S uvv − 2 v c S uv + v 2 c S u − α S u = 0
Since S u = 0 , this simplifies to
u c S uu + v c S uv = 1 2 (S uuu + S uvv ) Eq. 4
In a similar fashion, expanding Eq. 3 and using S v = 0 gives
u c S uv + v c S vv = 1 2 (S vvv + S vuu ) Eq. 5
Solving Eq. 4 and Eq. 5 simultaneously gives (u c , v c ).
original coordinate system is (x c , y c ) = (u c , v c ) + (x, y).
Then the center (x c , y c ) of the circle in the
To find the radius R, expand Eq. 1:
∑ [
u
2
i − 2 u i u c + u 2 c + vi 2 − 2 v i v c + vc 2 − α ] = 0
i
Using S u = S v = 0 again, we get
N ( u 2 c + v2 c − α ) + S uu + S vv = 0
Thus
α = u 2 c + v2 c + S uu + S vv
N
Eq. 6
and, of course, R = √ α.
See the next page for an example!