Locating a circle on the plane tangent to the unit circle (Differential Geometry)

Question

I was working on a file in Mathematica in preparation for a talk, and I have a quick question.I want to illustrate the idea of revolving a circle about another circle to make a torus. I would like to focus on plotting a circle (say $c_{1}$) and another circle (say $c_{2}$) that lies on plane perpendicular to $c_{1}$ in the direction of tangent vector of $c_{1}$.

I start with $x[t]:= \{\cos[t],\sin[t],0\}$ so computing $x'[t]:=\{-\sin[t],\cos[t],0\}$, but I specifically want to look at some $t$-value, say $\pi$. Now I take $\widehat{x'}:= \{0,-1,0\}$. Now I must find a vector perpendicular to $\widehat{x'}$, say $\widehat{p}:= \{-1,0,0\}$. Lastly I take the cross product of these to vectors, $\widehat{x'}$ $\times$ $\widehat{p}: = \{0,0,-1\}$.

And am done using, $x[t]+\cos[s]\{-1,0,0\}+\sin[s]\{0,0,-1\}$ for $s \in [0,2\pi]$.

I must have some false logic here because Mathematica isn't giving me anything.

Answer 1

The idea would be to first create the large circle: $$(R \sin[\theta], R \cos(\theta), 0)$$ Then create a small circle at a certain angle (say $\theta=0$): $$(0, R + r\cos[\phi], r \sin(\phi))$$ Now, multiply the small circle by a rotation matrix to position it at the angle you want. The Mathematica code below should do something like that.

R = 1; r = 0.2;  
RotationMat[ph_] := {{Cos[ph], -Sin[ph], 0}, {Sin[ph], Cos[ph], 0}, {0, 0, 1}}; 
MainCirc  = ParametricPlot3D[{R Sin[th], R Cos[th], 0}, {th, 0, 2 \[Pi]},         
              PlotRange -> {{-R - r, R + r}, {-R - r, R + r}, {-R - r, R + r}}]; 
smallCirc = ParametricPlot3D[RotationMat[70 \[Pi]/180].{0, R + r Cos[th], r Sin[th]}, {th, 0, 2 \[Pi]}]; 
Show[MainCirc, smallCirc]