Given a two dimensional Riemannian manifold with the property that the ratio of the circumference and the diameter is the same for all circles. What can be said about it? Does it have to be the Euclidean plane with the usual metric?
If $\dfrac{\mathrm {circumference}}{\mathrm {diameter}}$ is the same for all circles, does the surface have to be flat?
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Certainly not. It depends on the Space curvature.
For very "shallow" $C_2$ smooth patches of circle radius $r$ measured in geodesic polar co-ordinates,the circumference tends to $2 \pi r$. Here negligible or zero Gauss curvature $ K, $ what we call flat Euclidean geometry.
When we write $ |K|\, a^2 = 1 $ for "deeper" patches $ K$ is non-zero, positive or negative.The circumference /boundary length of closed loops are,
in elliptic geometry :
$$ 2 \pi a \;\sin (r/a) < 2 \pi r, $$
in hyperbolic geometry: $$ 2 \pi a \; \sinh (r/a) > 2 \pi r, $$
and in Euclidean geometry:
$$ 2 \pi r. $$
Example patches are spherical cap, Pringles chips saddle surface and flat disc respectively.
There are similar corresponding formulas for enclosed patch areas.
Gauss Bonnet Theorem and geodesic polar coordinates are the relevant study materials.
Let's agree that a "circle of center $p$ and radius $r > 0$" refers to "the image of the Euclidean circle of radius $r$ in the tangent plane at $p$ under the exponential map", and that the metric is of class $C^{2}$, i.e., smooth enough that the Gaussian curvature is continuous.
If the circumference of a circle of radius $r$ is $2\pi r$ merely for sufficiently small circles, the Gaussian curvature is zero. (Sketch of proof: If $K(p) \neq 0$, there exists a circle around $p$ inside which $K$ is bounded away from $0$; the Jacobi equation implies that radial geosedics through $p$ don't diverge like rays in Euclidean space, so the circle has the "wrong" circumference.)
Now we get into nitpicking:
With the preceding definition of "circles", the answer to the original question is "no"; all you can deduce is that the surface is flat (and perhaps complete, depending on the intent of the question).
If a "circle" is required to be an embedded curve, then "yes": In any complete flat surface other than the Euclidean plane, a circle of sufficiently large radius crosses itself.
It may be of interest to note that if $M$ is a cone with vertex $p$ (not of class $C^{2}$ at $p$), then circles centered at $p$ have circumference proportional to their radius, but the constant of proportionality is $\theta$, the incident angle at the vertex (and $\theta$ can be an arbitrary positive real number).