Can you make the circle into a vector space? I thought maybe use a set with elements of the form $e^{i\theta}$ but what field would the scalars be taken from? Multiplying $ke^{i\theta}$ with $k \in \mathbb{R}$ or $\mathbb{C}$ doesn't give you another point on the circle though, so what field is used?
For the unit circle in $\mathbb{R}^{2}$, call it $S = \{(x,y) \in \mathbb{R}^{2}:x^{2}+y^{2}=1\}$ is not a vector space. Indeed, the two vectors $(1,0)$ and $(-1,0)$ lie in $S$. We will show that the sum of these two vectors do not lie in $S$. From this, we can therefore conclude that $S$ is not a vetor space, as it will not be closed under addition. So since $(1,0)+(-1,0)=(0,0)$ and $(0,0) \notin S$, this finishes the proof.

Here is a question to contemplate. If one adjoins the zero vector $(0,0)$ to $S$, is $S \cup \{(0,0)\}$ a vector space?
For the unit circle in $\mathbb{R}^{2}$, call it $S = \{(x,y) \in \mathbb{R}^{2}:x^{2}+y^{2}=1\}$ is not a vector space. Indeed, the two vectors $(1,0)$ and $(-1,0)$ lie in $S$. We will show that the sum of these two vectors do not lie in $S$. From this, we can therefore conclude that $S$ is not a vetor space, as it will not be closed under addition. So since $(1,0)+(-1,0)=(0,0)$ and $(0,0) \notin S$, this finishes the proof.

Here is a question to contemplate. If one adjoins the zero vector $(0,0)$ to $S$, is $S \cup \{(0,0)\}$ a vector space?