The locus of a point $P$ such that there is a constant distance ratio $e=PA/PB$ between two fixed points $A$, $B$ in the plane is known to be the Apollonian Circle.
Correspondingly on a unit sphere what is the locus between points of long/lat $A(-L,0)$, $B(L,0)$ and $P(\operatorname{Long}P, \operatorname{Lat}P)$ along great circle arcs for a given $e$?
Is it is a small circle? If so what is its geodesic curvature $\kappa_g?$ The $\kappa_g$ should vanish for the prime meridian when $e=1.$
Constant $PA+PB$ "ellipses" were discussed here before ( @ achille hui ) ... Thanks in advance for literature if available, or, for a derivation/image if not available.
EDIT1:
Geodesic curvature $\kappa_g=1/R_g=\dfrac{\tan \gamma}{R}.$
Tangent length at any point is $R_g$ (red line segment) on the small circle and angle $\gamma$ is a dimensionless angle parameter showing how far the small circle has deviated from the great circle or the equator. $\gamma=0$ for the great circle geodesic.
Without loss of generality we may assume that $A=(-1,0,0)$ and $B=(\cos\alpha,\sin\alpha,0)$ and let $e>0$ be a real number. We want to find the locus of points $P=(x,y,z)$ on the unit sphere which has the property $\frac{PA}{PB}=e$. Here, $PA$ is the length of the shortest geodesic from $A$ to $P$. Similarly, for $PB$.
The arc $PA$ is measured by the angle $AOP$ say $\theta_1$ and the arc $PB$ is measured by the angle $BOP$ say $\theta_2$. And the points on the Apollonian circle satisfy the equation $\frac{\theta_1}{\theta_2}=e$ with the sphere.
Then from the standart $\Bbb{u}\cdot\Bbb{v}=|\Bbb{u}||\Bbb{v}|\cos\theta$ formula, we get $\theta_1=\arccos(-x)=\pi -\arccos(x)$ and $\theta_2=\arccos(x\cos\alpha+y\sin\alpha).$
Therefore, $P$ must line on the surface: $$\frac{\pi -\arccos(x)}{\arccos(x\cos\alpha+y\sin\alpha)}=e \tag{1}$$ And the Apollonian circle is the intersection of that surface with the unit sphere. As a special case when $\alpha =0$, that is when $B=(1,0,0)$, this surface is the plane $x=\cos\frac{\pi}{e+1}$, as we found earlier, and the Apollonian circle is the intersection of that plane with the sphere. And it is really a circle.
Let us be a little sneaky and take $P$ on the equator AB. Then $P=(\cos\theta,\sin\theta,0)$, $x=\cos\theta$ and $x\cos\alpha+y\sin\alpha=\cos(\theta-\alpha)$. And from the equation $(1)$ ve get $\theta=\frac{\pi+e\alpha}{e+1}$ and $$P=(\cos(\frac{\pi+e\alpha}{e+1}),\sin(\frac{\pi+e\alpha}{e+1}),0)$$ on shorther geodesic and similarly we find $$Q=(\cos(\frac{e\alpha -\pi}{e+1}),\sin(\frac{e\alpha -\pi}{e+1}),0)$$ on longer geodesic. And the plane passing through them and perpendicular to $xy$-plane is $$cos(\frac{e\alpha}{e+1})x+sin(\frac{e\alpha}{e+1})y=cos(\frac{\pi}{e+1}).\tag{2}$$ The Apollonian circle is the intersection of the plane $(2)$ and the unit sphere. Its radius is $R_e(A,B)=sin(\frac{\pi}{e+1})$ and its geodesic curvature is $\kappa_g=cot(\frac{\pi}{e+1})$.
But, the general formula is still open.