How to find center and radius of hand-drawn circle? You are given a set of points {(X1,Y1), (X2,Y2),...}  which represent a hand-drawn circle, so it's not perfect. You are asked to find the center and radius of this circle.

My intuition tells me this involves minimization, and I need to find the coefficients in a non-linear generalized least squares model. Am I not thinking of a simpler or more "proper" approach?
You can use a trick to make this a regular least-squares optimization, in an extension of the method for finding the center of a circle through three points. Draw a perpendicular bisector for every pair of points. If the points are all on a circle, they will all intersect at a common point; since it is an approximate circle, they will slightly miss each other. The squared distance to each line is a quadratic form, so the sum of the squared distances is also a quadratic form, and minimizing this is a least-squares problem. That's the center of the circle.

Given this, you can just find the radius as the mean of the distances of each point to the center.
For each point you are given, we assume that it satisfies the following relation:

$$
( x_i - x_0 )^2 + ( y_i - y_0)^2 = r^2.
$$

This can be written as

$$
\left[
\begin{array}{ccc}
1 &amp; -2x_i &amp; -2y_i
\end{array}
\right]
\left[
\begin{array}{c}
x_0^2 + y_0^2 -r^2 \\
x_0 \\
y_0
\end{array}
\right]
= -x_i^2 - y_i^2.
$$

Thus, using all of the $(x_i,y_i)$ pairs, we can set up an over-determined system $\mathbf{Ax}=\mathbf{b}$. The least squares solution $\bar{\mathbf{x}}$ to this system will give you your answer.