# Curvature of geodesic circles on surface with constant curvature

I am trying to solve the following exercise:

Prove that on a surface of constant curvature the geodesic circles have constant curvature.

"Constant curvature" in case of the surface I take to refer to the Gaussian curvature. Now, the geodesic curvature of a curve parameterized by arc length in orthogonal coordinates is given by

$$k_g(s) = \frac{1}{2 \sqrt{EG}} \left(G_u v'- E_v u' \right)+ \phi',$$

where $\cdot'$ denotes the derivative with respect to $s$, and $\phi$ is the angle the tangent of the curve makes with $x_u$.

Using geodesic polar coordinates (setting $u = \rho$ and $v = \theta$), a surface with constant Gaussian curvature $K$ satisfies

$$(\sqrt{G}_{\rho\rho}) + K \sqrt{G} = 0$$

Also, we get $E=1$, $F=0$, and a geodesic circle has the equation $\rho = \mathrm{const.}$ Therefore, the first equation above yields

$$k_g(s) = \frac{G_\rho \theta'}{2\sqrt{G}}$$

It seems to prove that $k_g$ is constant, you would have to show that its derivative is 0. I tried that, but the derivative gets rather ugly and I don't see how to proceed.

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