In Boyd & Vandenberghe's Convex Optimization, the affine hull of a subset $C \subset \mathbb R^n$ is defined 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 is claimed $\text{aff } U = \mathbb R^2$ 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.
Here is an easy analytic proof of $\text{aff}\,U=\mathbb{R}^2$:
Take any point $x\in\mathbb{R}^2$, which we will express in polar form: $x=\rho e^{i\varphi}$. Now, we can write $x$ as $$x = \theta (-e^{i\varphi})+(1-\theta)e^{i\varphi},$$ where $1-2\theta=\rho$. That is, the arbitrary point $x$ is an affine combination of two points $-e^{i\varphi}$ and $e^{i\varphi}$ of the unit circle, so $x\in \text{aff}\,U$.
Let me propose an analytical approach for better understanding the proof.
Lets consider the set $C = \{ x \in R² | x_1²+x_2²=1$}
The affine hull is $\text{aff} (C) = \left\{\theta_1 x_1 + \ldots +\theta_k x_k \mid x_1, \ldots x_k \in C, \theta_1 + \ldots \theta_k = 1 \right\}.$ So it is the set that contains all the affine combinations of $C$.
Lets consider any two opposite points in the circle $p_1, p_2 \in C$and its affine combination: $\theta p_1 + (1-\theta)p_2$ so that $\theta \in R$
Now lets write them in polar form for any $\alpha \in [0, 2\pi)$ (Notice that radius is 1): $p_1=(cos\alpha, sin\alpha)$, $p_2=(cos(\alpha+\pi),sin(\alpha+\pi))$
We operate the linear combination, keeping in mind that $sin(\alpha + \pi)= -sin(\alpha)$ and $cos(\alpha + \pi)= -cos(\alpha)$: $$(\theta cos\alpha + (1-\theta)(cos(\alpha+\pi), \theta sin\alpha + (1-\theta)(sin(\alpha+\pi))) = $$ $$= 2\theta-1(cos\alpha, sin\alpha) = $$
As $\theta$ and $2\theta-1$ are both constants in $R$ which depend of $\theta$, we simplify both in $\theta$, thus:
$$= \theta(cos\alpha, sin\alpha)$$
As $\theta$ varies, it traces a line that cross $p_1$ and $p_2$. As $\alpha$ varies, $p_1$ and $p_2$ vary also, so the lines are different. In conclusion, this affine combination traces out the entire $R²$ plane. (Here we should also proof that $\theta(cos\alpha, sin\alpha)$ is dense)
The central idea of this proof is that any point $p\in R^2$ is contained in the line generated by any two points in the circle.
