If you had a torus in $\mathbb{R}^3$ created by sweeping a circle of radius $1$ perpendicular to the path defined by the circle $x^2+y^2=1$ (or equivalently the parametrized version of the circle as $x=sin(\theta), y=cos(\theta), 0 \le \theta \le 2\pi$), then the point $(0,0,0)$ is contained in an infinite number of circles.

I believe you are asking about the outer hull of a swept surface defined by sweeping an outline/path object along another outline or path object.  As Michael Hardy pointed out, there is more than one way to specify a torus as the sweep of a circle.  In fact, there are four ways with the Villarceau circles.  Sometimes, specifying one of the non-obvious circles is a better way to solve certain problems (see https://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/ about the volume of napkin rings)

There's a great image of the non-obvious Villarceau circles at Benoît Kloeckner's web site:

http://www-fourier.ujf-grenoble.fr/~bkloeckn/images/villarceau.png

If the radius of curvature of the sweep-path becomes smaller than the radius of the circle being swept, then you get strangeness unless your meshing algorithm remembers to remove the extraneous mesh-points that are within the swept-volume's outer hull.

I am not sure I'm completely following your differentiation between *circled* vs. *hoop* surfaces, unless you mean that in a circled surface the swept circle is maintained perpendicular to the sweep path, whereas in a hoop surface, it is not.  Is that what you meant?

It's also possible to define a sweep surface where the radius of the swept circle also varies parametrically along the sweep path, and if you do that in two-dimensions, you can emulate the behaviour of an ink-pen or a marker where increasing the pressure (or slowing the speed of the pen) leads to a thicker weight line being drawn at certain regions of the "writing path".