Does the geometric locus of points bearing a fixed ratio between two other fixed points have a constant geodesic curvature?
Let $M$ be a 2-dimensional Riemannian manifold $ x,y∈M. $ Is the set of points ${z |\,d(z,x)/d(z,y)}= \lambda$ comparable to the Apollonian circle in analogy for geodesic deviation?
Can it be generalized to higher-dimensional Riemannian manifolds?
I was wanting to be able to frame a similar question from here
The best I can understand this question, it is asking the following:
Definition. Let $M$ be a Riemannian manifold, $\lambda>0$ is a number. Given two points $p,q\in M$ define the $\lambda$-bisector $B_\lambda(p,q)$ as the locus of points $x\in M$ such that $d(p,x)=\lambda d(q,x)$. (If $\lambda=1$, this is the ordinary bisector.)
Question. What is the shape of $B_\lambda(p,q)$? For instance, if $dim(M)=2$, is it true that for some $\lambda>0$, for all $p, q\in M$, $B_\lambda(p,q)$ is a constant curvature curve? (With curvature not prescribed in advance.)
I do not know what does it have to do with Apollonian packings though.
If this is indeed the question, then a direct computation shows that for the flat metric on $R^2$ all $\lambda$-bisectors are circles, hence, they all have constant curvature. But if $M$ is the hyperbolic plane or the sphere (constant curvature $-1$ and $1$ respectively) then for generic points $p, q$ and all values $\lambda\ne 1$, $\lambda$-bisectors do not have constant curvature. (This is again a direct computation.) I very much doubt that there is a non-flat Riemannian metric on a surface such that for some $\lambda$ all $\lambda$-bisectors have constant curvature. Proving this, would require writing a serious research paper.