I don't think that is quite the proof I was looking for. I'm not sure if I was clear. The question I had was how do we know the line from ##\dot b## to ##\dot p## ends at the tangent point ##\dot g##, and not another point on the large circle. If it is shown they are the same point, then the similar triangles completes the radius derivation.

I don't disbelieve they are the same point. The drawings line up and it FITS. Every drawing (and the helpful MP4 graphic) LOOKS right. I just don't quite see why.

@votingmachine Start with the diagram with ##\dot c,\dot p, \vec d## marked, and let ##\dot f,\dot g## be the centre and tangent point of the circle that is the solution. Now define ##\dot b## to be the point where the line from ##\dot g## through ##\dot p## crosses the circle again.

Then we can easily show that the triangles ##\dot g\dot b\dot c## and ##\dot g\dot p\dot f## are similar (since both are isosceles, and they share an angle), from which it follows that ##\dot b\dot c## is parallel to ##\dot p\dot f##, which is parallel to ##\vec d##.

So, to find this solution, all we need do is

find the point ##\dot b## where the line through ##\dot c## parallel to ##\vec d## but in reverse direction cuts the circle

find the point ##\dot g## where the line from ##\dot b## through \dot p## cuts the circle again. That is the tangent point

find the point ##\dot f## where the radius from ##\dot g## to ##\dot c## cuts the ray in direction ##\vec d## from ##\dot p##. That is the centre of the smaller circle.

which is the approach Ventrella described - albeit in slightly different words..

B–K1 is a diameter of the big circle. Thales' theorem gives the angle at G between B and K1 as 90°.
P–K2 is a diameter of the small circle. Thales' theorem gives the angle at G between P and K2 as 90°.
Since the diameters B–K1 and P–K2 are parallel, and the angles at G are the same, the point K2 must lie on the small circle and they must be similar triangles.