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I am not sure how to do this. $S$ includes singletons too (radius $0$). I have tried mapping a circle to the pair consisting of its center point and its radius, but it isn't injective.

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    I have not worked out the details yet, but it might be possible to give a structure to all circles in the plane, and then use stereo graphic projection from the north and south poles of the sphere? – 2017-01-30
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    That is probably possible. How do I give a structure to all circles in the plane? – 2017-01-30
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    In the plane it seems like your idea of using the center point and its radius would work – 2017-01-30
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    How am I going to handle circles which go through the poles? – 2017-01-30
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    You're right, could be an issue...I will think some more. *Possibly* doing stereographic projections from several natural points ( I.e top, bottom, front, back, left, right ) so that no circle passes thru each. – 2017-01-30
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    What do you mean by "the set of all circles"? What's the topology there? – 2017-01-30
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    @freakish I am not given any topology. I am supposed to pick an appropriate, "natural" topology. – 2017-01-30
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    Without topology the task is trivial. The set is equinumerous with reals. So the bijection induces manifold's structure. – 2017-01-30
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    But Im pretty sure you have some topology on the set. For example induced by Hausdorff distance. That is more interesting. – 2017-01-30

3 Answers 3

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A circle is just the intersection of the sphere with a plane. The case of point is the case of tangent planes. The set of affine planes in $R^3$ is naturally a manifold of dimension 3 (in fact it is the tautological line bundel over $RP^2$, and your manifold appears as a closed submanifold with boundary the sphere $S^2$ in it. It is sliglty easier to describe the set of oriented circles, which are intersection of oriented planes with the sphere. An oriented plane is given by a unit (normal) vector u, and a parameter $t\in R$ with the equation $=t$, where $X=(x,y,z)$ is the generic point. Note that this plane meets the sphere iff $t\leq 1$, so the set of oriented circle is just $S^2\times [-1,1]$. For the case of unoriented circles, you can check that is is the subset of the tautological line bundle over $P^2$ made up with points at the distance $\leq 1$ to the origin. Its boundary is the 2-sphere as expected.

Note the beautiff action of $SO(3)$ on this manifold wih is transitive on each family of circle of given radius, proving that this set is a sphere, unless the radius is 0.

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If you use stereographic projection $(X,Y,Z) \mapsto (x,y)$ of the sphere from its North Pole $N$ onto its equatorial plane (which is known to be a space inversion) as advised by @ThomasGrubb,

$$\tag{0}\begin{cases}x=\dfrac{2X}{1+X^2+Y^2}\\y=\dfrac{2Y}{1+X^2+Y^2}\end{cases}$$

The image of a circle not passing through $N$ is known to be a circle (https://people.maths.ox.ac.uk/earl/G2-lecture5.pdf); the image of a circle passing through $N$ is a straight line. We can consider straight lines as "generalized circles" (intuitively with "infinite radius"... with quotes.)

The union of ordinary and generalized circles can be parameterized with $4$ coordinates $a,b,c,d$ by considering the general equation:

$$d(x^2+y^2)-2ax -2by+c=0$$

$$\text{if} \ \ \begin{cases}d\neq0 \text{: circle with center} \ (a,b) \ \text{and radius} \ R=\sqrt{a^2+b^2-c},\\ d=0 \text{: straight line}\end{cases}$$

In fact, this representation accounts for the projective nature of this space.

If one desires to takes into account the sign of $d$, we get Laguerre's oriented circles' geometry.

There is a classical quadratic form, providing a riemannian structure, on this space (see the space of spheres in chapter XX of (http://link.springer.com/content/pdf/10.1007%2F978-1-4757-1836-2_20.pdf):

$$\tag{2}q(a,b,c,d)= a^2+b^2-cd$$

Explanation : when restricted to the circles, this expression becomes, using (1):

$$\tag{3}q(a,b,c,1)= a^2+b^2-c=R^2$$

The fact that $q$ accounts for the square of a radius, i.e., the square of a size shows that $q$ is a rather "natural" choice.

Remark: (1) can be written:

$$q(a,b,c,d)= a^2+b^2-cd=a^2+b^2+\frac{1}{4}\left(c-d\right)^2-\frac{1}{4}\left(c+d\right)^2$$

thus with a $(+,+,+,-)$ signature, which is a Minkovski-type structure, such as one finds it in space-time geometry.

It suffices now to "lift" quadratic form (2) onto the sphere.

Note that we have worked on the sphere deprived of point $N.$ This is a map (in the manifold sense of the word). We need a second map, in order that this atlas of two maps provides a (riemannian) manifold structure to the sphere.

An evident choice is stereographic projection from the south pole $S$.

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Here is my favorite way to see this. Consider $R^3$ as an affine chart in $RP^3$, the real-projective space. Let $B\subset RP^3$ denotes the closed unit ball, with the boundary sphere $S^2$. For any point $p\in RP^3 - B$ consider the conic which is the union of lines through $p$ tangent to $S^2$. The set of points of tangency is a circle. Conversely, every circle in $S^2$ appears this way. One way to see this is to find a projective transformation preserving $B$ and sending the given circle $C$ to a great circle $C'\subset S^2$. Then for $C'$ the relevant conic is the cylinder through $C$ with the axis $L$ passing through the center of $B$ (plus the point $p$ in $RP^3 - R^3$ represented by the line $L$). Thus, with any reasonable topological structure on the set of circles of positive radius in $S^2$, it is homeomorphic to $RP^3 - B$, which is a manifold, being an open subset of $RP^3$. Now, if you allow circles of zero radius,you have to add to $RP^3 - B$ the boundary sphere $S^2$: I a sequence of points $p_n\in RP^3 -B$ converges to a point $p\in S^2$, the corresponding circles $C_n\subset S^2$ converge to $p$ as well. Hence, your space $S$ is homeomorphic to $RP^3 - int(B)$, which is a smooth manifold with boundary.