TITLE: Curvature of geodesic circles on surface with constant curvature
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: If you have textbook, Differential Geometry of Curves and Surfaces, Do CARMO,
Then See p.289, the part of Theorem by Minding. You will get $E=1,F=0,G=constant,G_ρ=constant$ in geodesic circles. (i.e, ρ=constant)
And see p.254, the part of Theorem by Liouville, curvature of curve ρ=constant,
$k_{g2}=\large\frac {G_ρ}{2G\sqrt{E}}$. So k is constant.