Why is the affine hull of the unit circle $\mathbb R^2$?

In Boyd's "Convex Optimization" it defines the affine hull of a subset $C$ of $\mathbb R^n$ as

$$\text{aff} C = \left\{\theta_1 x_1 + \ldots +\theta_k x_k \mid x_1, \ldots x_k \in C, \theta_1 + \ldots \theta_k = 1 \right\}.$$ Then, it claims $\text{aff } U = \mathbb R^2$ if $U$ is the unit circle. Why is this? Isn't any arc (or convex subset of the circle) entirely contained in the circle? I would think $\text{aff } U = U$ if $U$ is the unit circle.

I guess I know your confusion. The point is that $\theta_i$ is not constrained to be greater than zero. So now you may understand why three non-collinear points will fill the whole $\Bbb R^2$.

To have an answer recorded as such, I'll add a few words. If $C$ contains the origin, then the affine hull is the same as linear span, since we can include $0$ with any coefficient we want. Also, translating $C$ by a vector translates its affine hull by the same vector. Thus, we can find the affine hull by moving the coordinate system so that the origin lies in $C$, and then taking the linear span. This shows at once that the affine hull of any three non-collinear points in the plane is the entire plane.

Take any point in $\mathbb R$. We can always draw a line through it which passes through two points in the circle. That means it lies on the same line that passes through those points on the unit circle.

That's the definition of Affine Hull. Please rectify me if I'm wrong.

The unit circle $U$ is the set of all points (x,y) such that $x^2 + y^2 = 1$. So, if we take the affine hull of $U$, we will generate $\mathbb{R}^2$ since there exists at least 3 non-collinear points in $U$.

In particular, all we need to look at is a three element subset of the unit circle where the three points do not all lie along a line in order to generate $\mathbb{R}^2$ as a set of affine combinations.

If you are still confused by my answer and LVK's answer, you may want to review the definition of dimension and affine combination.