Points at twice the distance from (-1, 0) that they are from (1, 0) in hyperbolic geometry [closed]

In answer to the question Demystifying complex numbers, Charles Matthews suggests "finding the points at twice the distance from (-1, 0) that they are from (1, 0)." as a motivation for complex numbers.

Suppose you want to find these points in hyperbolic geometry instead of euclidean geometry. If this can be done with vectors or complex numbers in R^2, then I reckon it could be done with gyrovectors or gyro-complex numbers in the hyperbolic plane, but if you don't use gyro-algebra then how would you find (or describe) the points at twice the distance from (-1, 0) that they are from (1, 0)?

(Defining coordinates in hyperbolic geometry can be done with gyro-algebra, but without it just assume the origin is an arbitrary point, and that (-1,0) is a point of distance 1 from the origin and (1,0) is a point of distance 1 from the origin in the opposite direction.)

-

closed as too localized by Robin Chapman, Yemon Choi, S. Carnahan♦Jul 2 '10 at 20:48

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

While Robin Chapman has given a nice answer, I think the question seems a bit basic/localized for MO (unless I have misunderstood something) – Yemon Choi Jul 2 '10 at 19:14
This is not so much a real question as another advertisement for certain constructions by one Abraham Ungar, see ndsu.edu/pubweb/~ungar and ndsu.edu/pubweb/~ungar/publications.html – Will Jagy Jul 2 '10 at 20:19

In the upper half plane model the distance satisfies $$d(a+bi,c+di)=\cosh^{-1}\frac{(a-c)^2+b^2+d^2}{2bd}.$$ As each line in the upper half plane can be transformed into the imaginary axis, we can take our two points to lie on this axis. So let $u>v>0$ and seek the $z=x+yi$ with $$d(z,ui)=2d(z,vi).$$ Using the identity $$\cosh 2t=2\cosh^2t-1$$ we get $$\frac{x^2+y^2+u^2}{2yu}=\frac{(x^2+y^2+v^2)^2}{2y^2v^2}-1$$ that is $$v^2y(x^2+(y+u)^2)=u(x^2+y^2+v^2)^2,$$ a quartic curve.