# Geodesic on half-plane determined by tangent vector

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

Let a geodesic for the upper-half plane have the center $(a,0)$ and radius $r$. Thus it has the equation
$$(x-a)^2 + y^2 = r^2.$$
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.