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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 at $t=0$: $\dot r > 0$ 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 intersection. Likewise, if at $t=0$: $\dot r < 0$ 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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. It also suggests some twistor-like semantics for all this thing, namely, that null geodesics in the circles’ space are families of tangent circles on the plane. Ī hope this “elementary” approach can be helpful for somebody.

Disclaimer: my original posting made wrong statements about geometry, but Ī opted to preserve and fix it because a different approach ultimately resulted in similar conclusions than 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 look on symmetries acting 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 at $t=0$: $\dot r > 0$ 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 intersection. Likewise, if at $t=0$: $\dot r < 0$ 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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. It also suggests some twistor-like semantics for all this thing, namely, that null geodesics in the circles’ space are families of tangent circles on the plane. Ī hope this “elementary” approach can be helpful for somebody.

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 at $t=0$: $\dot r > 0$ 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 at $t=0$: $\dot r < 0$ 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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. Ī hope it can be helpful for somebody.

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 at $t=0$: $\dot r > 0$ 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 intersection. Likewise, if at $t=0$: $\dot r < 0$ 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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. It also suggests some twistor-like semantics for all this thing, namely, that null geodesics in the circles’ space are families of tangent circles on the plane. Ī hope this “elementary” approach can be helpful for somebody.

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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$.

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 at $t=0$: $\dot r > 0$ 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 at $t=0$: $\dot r < 0$ 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 $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. Ī hope it can be helpful for somebody.