Let $\gamma: \mathbb{R} \to \mathbb{R}^2$ be a simple closed curve with period $T>0$. Let $\Delta$ be its circumscribed disk, show that the bounded component of $\mathbb{R}^2 \setminus \gamma(\mathbb{R})$ is contained in $\Delta$.
This is what I have done so far:
Since $\gamma$ is a simple closed curve by Jordan's theorem $\gamma$ divides $\mathbb{R}^2$ into two components: int$(\gamma)$, which is the bounded component, and ext $(\gamma)$. Therefore I need to show that int$(\gamma) \subset \Delta$. Let's suppose that $int(\gamma) \nsubseteq \Delta$, that is, that there exists a point $q \in int(\gamma)$ such that $p\notin \Delta$ (where $r$ is the radius of $\Delta$ and $p$ is its centre). Thus $d(p, q) > r$. I need to show now that this implies that a piece of $\gamma$ is not contained in $\Delta$, but I don't know how to. Could anyone help me?
Let $B = \mathbb R^2 \setminus \Delta$ the exterior of the disk $\Delta$. Write $\Gamma = \gamma(\mathbb R)$. Since $\Gamma \subset \Delta$, we see that $B \cap \Gamma = \emptyset$. $B$ is connected and must therefore be contained in one the two connected components $\operatorname {int} \gamma$, $\operatorname {ext} \gamma$ of $\mathbb R^2 \setminus \Gamma$. Since $B$ is unbounded, we see that $B \subset \operatorname {ext} \gamma$. Thus $$\operatorname {int} \gamma \subset \operatorname {int} \gamma \cup \Gamma = \mathbb R^2 \setminus \operatorname {ext} \gamma \subset \mathbb R^2 \setminus B = \Delta .$$