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The upper-half plane $\mathbb H$ carries a hyperbolic metric and the geodesics are semicircles with base on the real line. We consider oriented geodesics. Let $x \in \mathbb H$ and let $v$ be a unit tangent vector at $v$. How to prove the following statement:

There exists a unique geodesic $\gamma$ on $\mathbb H$ such that $\gamma^\prime(0) = v$.

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This fact is true on any Riemannian manifold, though $\gamma$ is only neccesarily defined for small $t$ values. On a complete Riemannian manifold (like $\mathbf{H}$), $\gamma$ is defined for all time. –  Jason DeVito Nov 23 '10 at 1:34
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Here is an attempt at solution without getting into the intricacies of Riemannian geometry.

Let a geodesic for the upper-half plane have the center $(a,0)$ and radius $r$. Thus it has the equation

$$(x-a)^2 + y^2 = r^2.$$

Specifying a unit tangent vector at a point is a choice of a direction there, which might be given by a directed line passing through that point. Your requirement translates to the assertion that the slope of the circle at this point agrees with that of the directed line corresponding to the unit vector.

We have two unknowns in this situation, viz., $a$ and $r$. Partial differentiation of the circle equation to get the $x$ and $y$ components of slope would give two linear equations, for which there would be a unique solution.

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