A moment's thought (or some careful examination of maps) reveals that any circle drawn on a globe is in fact a circle in real life. The same claim holds for circles drawn on flat surfaces (obviously).
Are these the only examples?
Formally, let $M$ be any Riemannian surface, (isometrically) embedded in $\mathbb{R}^n$. Moreover, suppose that any metric circle $C$ (i.e., a locus of points with constant geodesic distance from some fixed point) is in fact a circle in $\mathbb{R}^n$. Then must $M$ have constant nonnegative Gaussian curvature?
Yes!
First, note that such a surface must have nonnegative curvature. For: fix a point $p$. Working in geodesic polar coordinates about $p$, the group $\mathrm{SO}(\mathbb{R}^2)\cong S^1$ acts on any connected surface by rotating each metric circle. Since our metric circles are genuine circles, the resulting action is an isometry on some geodesic ball around $p$. But rotation by $90^\circ$ interchanges the principal curvatures; thus both curvatures are equal. Since the Gaussian curvature is the product of the two curvatures, it is a perfect square and so nonnegative.
Second, since genuine circles have constant geodesic curvature, this Q&A thread proves that $M$ must have constant curvature.