Tangent length between dragged point $P$ and circle lead (instead of straight line of Tractrix) is constant. What is the parameterization of such curves?
This is an attempt to "bend" the tractrix meridians replacing line of its symmetry by a circular asymptote.. while still maintaining contant tangent length property upto its new bent circle asymptote of radius $R$.
EDIT1
For time being a numerical solution is attempted in self answering mode reporting result obtained so far. It has an upper bent tractrix and a lower tractrix. The lower case results in a shape similar to a cardoid (or spherically reflected rays envelope caustic ) which is briefly presented.
$$ R^2 = r^2 + a^2 - 2 r a \cos (\pi-\psi) \tag1$$
Arc differentiation gives
$$\tan \psi \cdot \psi '=\left( \frac{1}{a}+\frac{\cos \psi}{r} \right) \tag2 $$
Differential geometric relations in a triangle
$$ r^{'} = \cos \psi; \quad \theta^{'}= \sin \psi/r \tag3 $$
This leads to a property of the bent tractrix:
$$ r\cdot (r+ 2 a \cos \psi) = (R^2-a^2) = R_g^2 \tag4 $$
where the radius of curvature $CP$ increases rapidly to $R_g $
Constants chosen for drawing are $a=0.6, R=1$.
Bent Tractrix in blue is from CAS. Red tangents are drawn by hand.
The most remarkable property of the Bent Tractrix is its relation to the following well-known property of the Circle with its external or internal power property$ (T^2>0, T^2<0) $ as product of segments from origin is $ r \cdot (r+ 2 a \sin \psi) = T^2 $ that I have not yet fully figured out.. that auxiliary circle has not yet been drawn.
To find center of curvature $C$ and radius of curvature $R_g$ we have the relation
$$\tan {\angle CTP}= \frac{a}{R_g} = \frac{ a+ r \cos \psi}{ r \sin \psi }$$
All comments appreciated.
The tractrix of a circle is another (smaller) circle whose area is pi.a² smaller than the original; where "a" is the length of the "leash". Thus, the tractor of a circle does NOT have the original circle as the asymptote.