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.)

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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 2010 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 2010 at 20:19

closed as too localized by Robin Chapman, Yemon Choi, S. Carnahan♦Jul 2 2010 at 20:48

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.