Breakdown of the Least Squares Method to Solve for a Circle Given Points (Numerically)

We aim to fit a circle to the given set of points $(x_i, y_i, z_i)$, but since all the $z_i = 1$, the problem simplifies to fitting a circle in the $xy$-plane.

The general equation of a circle in 2D is:

$(x - a)^2 + (y - b)^2 = r^2$

Expanding this equation:

$x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2$

Rewriting it as:

$x^2 + y^2 - 2ax - 2by = r^2 - a^2 - b^2$

Let $D = -2a$, $E = -2b$, and $F = r^2 - a^2 - b^2$, then the equation simplifies to:

$x^2 + y^2 + Dx + Ey + F = 0$

Step 1: Setting Up the System

Given the four points:

1. $(2,0)$
2. $(0,1)$
3. $(-1,0)$
4. $(0,-1)$

For each point, we substitute into the equation:

$x^2 + y^2 + Dx + Ey + F = 0$

$x$	$y$	$x^2 + y^2$	Equation
2	0	4	$4 + 2D + 0E + F = 0$
0	1	1	$1 + 0D + 1E + F = 0$
-1	0	1	$1 - D + 0E + F = 0$
0	-1	1	$1 + 0D - E + F = 0$

Step 2: Forming the Linear System

From the above equations:

1. $4 + 2D + F = 0$
2. $1 + E + F = 0$
3. $1 - D + F = 0$
4. $1 - E + F = 0$

Rearrange into a matrix form:

2 & 0 & 1 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} D \\ E \\ F \end{bmatrix} = \begin{bmatrix} -4 \\ -1 \\ -1 \\ -1 \end{bmatrix}$$ #### Step 3: Solving for $$D, E, F$$ Using elimination: 1. From equations 2 and 4: - $$1 + E + F = 0$$ - $$1 - E + F = 0$$ Adding both: $$2 + 2F = 0 \Rightarrow F = -1$$ 2. Solve for $$E$$: $$1 + E - 1 = 0 \Rightarrow E = 0$$ 3. Solve for $$D$$: $$1 - D - 1 = 0 \Rightarrow D = 0$$ 4. Solve for $$a, b, r^2$$: Since $$D = -2a = 0 \Rightarrow a = 0$$ Since $$E = -2b = 0 \Rightarrow b = 0$$ Since $$F = r^2 - a^2 - b^2 = -1 \Rightarrow r^2 = 1$$ $$r = 1$$ ### Conclusion: The least squares circle has: $$\text{Center } (0,0) \quad \text{and radius } r = 1.$$ Would you like additional numerical details or explanations? --- ### Related Questions: 1. How do we handle noisy data when fitting a circle using least squares? 2. What if the data points do not lie exactly on a circle? How does the method adjust? 3. How can we modify the least squares method to fit an ellipse instead? 4. What are the advantages of using the algebraic form over the geometric form in least squares fitting? 5. How can we extend this method to 3D sphere fitting? ### **Tip:** If your data is noisy, you can use **total least squares (TLS)** instead of ordinary least squares (OLS) to minimize perpendicular errors rather than just algebraic ones!