Disclaimer: original posting made wrong statements about geometry, but Ī opted to preserve and fix it because a different approach ultimately resulted in similar conclusions that of other answers.
Addressing the second part (Euclidean) of the question.
There is very obvious parametrization of (Euclidean) circles: each has the centre (a point on the plane) and radius (a number), so as a smooth manifold their space is E2 × (0, +∞) (if we exclude all degenerate cases). Let’s use respective real coordinates $(x,y,r)$. To address the question of its geometry one first should determine which symmetry group act on it.
Which transformations of Euclidean plane preserve all circles? Obviously, similarity transformations (although not only). The next interesting question is a nature of their action, and the following observation will help us:
Any similarity transformation preserves quadratic form (or symmetric 0,2-tensor) fields $\frac{1}{r^2}(dx^2 + dy^2)$ and $\frac{1}{r^2}dr^2$ on the space of circles.
How could we guess that they should combine to a pseudo-Riemannian structure? There can be at least two arguments. One, “global”, is to consider inversion about a circle (that preserves geometry of circles) and look what happened to quadratic forms. Another, “local”, is to consider paths in the space of circles. Let’s consider such smooth path $φ(t)$ that $\dot φ(0) ≠ 0$, and look whether will $φ(t)$ for small $t$ intersect $φ(0)$. If $\dot r > 0$ and $(\dot r)^2 > (\dot x)^2 + (\dot y)^2$ ($r$, $x$, and $y$ refer to respective coordinates of $φ(t)$), then for some $ε>0$ circles $φ(t)$ for $0<t<ε$ will lie outside $φ(0)$, without intersections. Likewise, if $\dot r < 0$ and $(\dot r)^2 > (\dot x)^2 + (\dot y)^2$, then for some $ε>0$ circles $φ(t)$ for $0<t<ε$ will lie inside $φ(0)$. It suggests that the correct quadratic form must be conformally equivalent to $dr^2 - dx^2 - dy^2$, that (together with preservation requirements) yields $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$.