Ways of understanding the complement of a circle in $\mathbb{R}^3$ is homeomorphic to a solid torus minus a point

Question

I read that the complement of a circle in 3d space is a solid torus minus a point. The only way i could understand it is by thinking $\mathbb{R}^3$ as $S^3\setminus(0,0,0,1)$ via the stereographic projection, then from the Hopf map $p:S^3\to S^2$ we know that the preimage of a point $x$ in $S^2$ is a circle and the preimage of $S^2\setminus \{x\}$ is a solid torus with the point $(0,0,0,1)$ so by removing the point we have the solution to the statement.

I wanted to ask:

• Is the above solution correct or even a good way to think about it?
• What other ways are there to understand or even visualize this? For example we can show that the 3-sphere is the union of two solid tori $$ S^3=\partial D^4=\partial(D^2\times D^2)=(\partial D^2\times \overline{D^2})\cup(\overline{D^2}\times\partial D^2)=(S^1\times\overline{D}^2)\cup(\overline{D}^2\times S^1)$$ is there a similar way to show that the 3-sphere minus a circle is a solid tori?