DE of curvilinear triangle sides on sphere

Question

Can we find a common differential equation of three lines on a sphere which form a curvilinear scalene triangle with internal angle sum $=\pi ?$ (I.e., no spherical excess.)

By common differential equation we can ensure the same character of curvature for its sides.

( A special case of constant geodesic curvature i.e., circles of a particular radius can always be chosen to cut meridians symmetrically at $ 30^{\circ}$ forming an equilateral triangle of angle sum $\pi $ but a general situation is sought).

The question troubled me for quite some time ...

Answer 1

Earlier answered by achille hui to same question from me.

Three concurrent small circles on a sphere intersect to enclose a curvilinear triangle of three arcs with sum of vertex angles $\pi$ , or zero "spherical excess".

This can be shown in stereographic projection where point of concurrency is the North Pole... which may be shifted to any other point on the sphere.

EDIT1/2

The picture shows original lines and their three small circles ( Blue,Green,Magenta) passing through North Pole and intersection points determine vertices of a triangle with angle sum $\pi$

A differential equation of small circles is requested. For time being top view is made on inversion basis trig inverting respect to $M$ the bowl or mirror of inversion, obeying relations:

$$ NP\cdot NQ= NS^2= NR^2= NM^2= 4a^2 $$

$$ 1/u^2= 1/\rho^2+ 1/(4a^2),\, u= \frac{2 a \rho}{\sqrt{4a^2+\rho^2} },\, r_1= u \sin\alpha=\frac{4 a^2 \rho}{4a^2+\rho^2 }, \, r=r_1\sec (\theta+ \beta) $$

where $\beta$ is longitude.