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

October 24, 2006 10:22 am MDT Page 1 of 3
Least­Squares Circle Fit
R. Bullock
Given a finite set of points in R 2 , say {(x i , y i ) | 0 ≤ i < N }, we want to find the circle that “best”
(in a least­squares sense) fits the points. Define
x = 1 ∑
x i and y = 1 ∑
y i
N
N
i
and let u i = x i − x, v i = y i − y for 0 ≤ i < N. We solve the problem first in (u, v) coordinates,
and then transform back to (x, y).
Let the circle have center (u c , v c ) and radius R. We want to minimize S = ∑ i (g(u i, v i )) 2 , where
g(u, v) = (u − u c ) 2 + (v − v c ) 2 − α, and where α = R 2 . To do that, we differentiate S(α, u c , v c ).
i
∂S
∂α = 2 ∑ i
= −2 ∑ i
g(u i , v i ) ∂g
∂α (u i, v i )
g(u i , v i )
Thus ∂S/∂α = 0 iff
∑
g(u i , v i ) = 0 Eq. 1
i
Continuing, we have
∂S
∂u c
= 2 ∑ i
= 2 ∑ i
g(u i , v i ) ∂g
∂u c
(u i , v i )
g(u i , v i )2(u i − u c )(−1)
= −4 ∑ i
(u i − u c )g(u i , v i )
= −4 ∑ i
u i g(u i , v i ) + 4 u c
Thus, in the presence of Eq. 1, ∂S/∂u c = 0 holds iff
∑
g(u i , v i )
i
} {{ }
= 0 by Eq. 1
∑
u i g(u i , v i ) = 0 Eq. 2
i
Similarly, requiring ∂S/∂v c = 0 gives
∑
v i g(u i , v i ) = 0 Eq. 3
i

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!

October 24, 2006 10:22 am MDT Page 3 of 3
Example : Let’s take a few points from the parabola y = x 2 and fit a circle to them. Here’s a table
giving the points used:
i x i y i u i v i
0 0.000 0.000 -1.500 -3.250
1 0.500 0.250 -1.000 -3.000
2 1.000 1.000 -0.500 -2.250
3 1.500 2.250 0.000 -1.000
4 2.000 4.000 0.500 0.750
5 2.500 6.250 1.000 3.000
6 3.000 9.000 1.500 5.750
Here we have N = 7, x = 1.5, and y = 3.25. Also, S uu = 7, S uv = 21, S vv = 68.25, S uuu = 0,
S vvv = 143.81, S uvv = 31.5, S vuu = 5.25. Thus (using Eq. 4 and Eq. 5) we have the following 2 × 2
linear system for (u c , v c ):
[ 7 21
21 68.25
] [
uc
v c
]
=
[ ] 15.75
74.531
Solving this system gives (u c , v c ) = (−13.339, 5.1964), and thus (x c , y c ) = (−11.839, 8.4464). Substituting
these values into Eq. 6 gives α = 215.69, and hence R = 14.686. A plot of this example appears
below.