@vv  No, it's not compact, it's professional☺.

@vv  

Thank you for your participation. Yes, it can be proved both geometrically and analytically.
This example was made specifically because of the animation. (I really like moving pictures.)

@C_R  
Well, you can try to evaluate the quality of the translation on one text from the MapleSoft application center. By the way, there is also a description of the Draghilev method. Translation from Russian to English online Google translator.

@C_R 
In other words, you don't know about the existence of online translators?

@sija 
 

I will show you my example with the rolling of the original ellipsoid
x1^2 + 5*x2^2 + 4*x3^2 - 0.25 = 0
on the transcendental surface
x3 = - 0.25*x1^2*sin(x1) - 0.25*sin(1.5*x2).

Of course, the greater the accuracy, the more time is required. To get animation on an old computer, we choose the optimal option. The equation of the ellipsoid at each point is printed on the graph.
for_an_ellipsoid.mw

(I have many different examples on this topic on the forum. I don't provide text for every example, because it's too cumbersome, but I try to show the basic program.)

@C_R  
In this case sqrt(b[5] ^2+ b[6]^2) is for the working point to move uniformly along the circle and also to make it easier to get the integration interval. But this has nothing to do with Draghilev's method. If you want, I can give you a list of literature in Russian that preceded the method, well, the method itself in Russian.
The author died long ago, and I do this as a hobby. I am not a professional. But I repeat, I will answer any questions about the Method.

@sija  I don't know, it seems to me that 2d examples are not very different from 3d ones. I would advise to immediately work on rolling the surface, or rather, simultaneously with the curve.

@sija 
Very good, +.
I really like pictures like this. And I myself sometimes have fun this way.   
Are there any plans for examples of rolling in 3D and plans for rolling of surfaces too?

@C_R  

I didn't see your message because there was no request to me. I came across this place completely by accident. Click "Reply" and then I will know because the flag in the upper right corner of the page will turn red for me.
b[i] is a way to make Maple do natural parameterization when I use internal procedures. Such actions are much easier to perform if you write the code for solving the ODE yourself.

Yes, that's right: we get successive solutions and do this based on solving the Cauchy problem. That is, we solve a system of nonlinear equations using a numerical solution of the ODE.
In this case, it is the Draghilev method.

@vv  OK, thank you. 

@vv  
I don't know, maybe it's all about the accuracy of the ODE solution, but in all positions of the circle, the fsolve procedure shows, in addition to the point of tangency, the presence of intersection points with the transcendental curve.

@vv 
I don't agree with you, because near your solution, in addition to the point of tangency, there immediately appears a point of intersection of the circle with the curve. After all, I suggested an approach to finding a solution, or not?
I would like it without intersection and with one point of contact.

@vv 
As always, everything is great with you, +. Only I already said that I know this solution. I wanted to get exactly the case when the point of tangency is unique on each curve. 

@mmcdara   @C_R  
What was meant is that each curve touches the circle at only one point. But, of course, formally this is a solution.
Your solution is contained in the set of solutions that are presented in the animation. Here is approximately that same frame.
(The animation itself is done with a rough step.)

It is clear that there may be moments when the circle formally touches both curves, but simultaneously intersects one of them. And in this case the same situation arises.
Let's look at the lower right corner of the overall graph.

To find a solution to this problem, we can, for example, do the following: starting to construct inscribed circles, watch the moment when the inscribed circle crosses the transcendental curve. And as soon as the intersection occurs, fix the previous position of the circle. If you do this with a sufficiently small integration step, then you can talk about a good approximation to the solution.
The coordinates of the inscribed circle are approximately [0.584307201828654, -1.27793596947425], and the maximum radius is somewhere around 0.656227713070106.
IN_EL_SIN_1.mw