Synthetic geometry: stereographic projections of $\mathbb{C}$ on Riemann sphere $\Sigma$ are inversions in sphere $K$ centered on $\infty$ of $\Sigma$

Question

The complete statement is the following:

Show that if $K$ is the sphere of radius $\sqrt{2}$ centered at the north pole ($N=\infty$) of the Riemann sphere $\Sigma$ s.t. $K$ intersects $\Sigma$ about the unit circle $C$ of $\mathbb{C}$:

Then, for a point $a$ on $\mathbb{C}$ and its stereographic projection $\hat{a}$ on $\Sigma$, $\hat{a}$ is the inversion in $K$ of $a$, and $a$ is the inversion in $K$ of $\hat{a}$. Notice that, on the image, the unit circle $C$ is represented by the line from $-1$ to $+1$.

I'm looking for a proof on this 2D graph with circles, using no algebraic property, i.e. pure synthetic geometry. Naturally this will trivially generalize to spheres.

Here's what I've done up to now. First, remember the geometrical definition of circle inversion of $a$ in $K$: Where $qI1\hat{a}$ and $qI2\hat{a}$ are right angles, and $I1\hat{a}, I2\hat{a}$ are tangents to $K$.

The theorem is very obvious if one constructs the two following machines where either $a$ or $\hat{a}$ is fixed on the line $\mathbb{C}$:

CASE 1. $a$ on the complex plane and within $C$:

CASE 2. $a$ off the complex plane:

The problem is, of course, to produce the geometrical proof that these constructions do hold the property of necessarily making $\hat{a}$ the inversion in $K$ of $a$, and $a$ the inversion in $K$ of $\hat{a}$, no matter what point on $\mathbb{C}$ we might choose.

This problem is from Needham's Visual Complex Analysis, p. 142-143; the author offers an algebraic solution while I'm looking for a geometrical one.

Answer 1

We can use the fact that the sphere $\Sigma$ passes through the point $N$, which is the center of the sphere of inversion $K$. It is a geometric theorem that the inversion of some sphere passing through the center of inversion is mapped to a plane. Every point on the unit circle $C$ intersects the sphere $K$, and hence they are all mapped to themselves. The unique plane intersecting $C$ is exactly $\mathbb{C}$, hence when $\Sigma$ is inverted through $K$, it is mapped exactly to $\mathbb{C}$.

More explicitly, take any point $a\neq N$ on the sphere $\Sigma$. Where $S$ is the south pole of $\Sigma$, since $SN$ forms a diameter of $\Sigma$, it follows from Thales's theorem that $\angle NaS$ is a right angle. By inverting through $K$ we have $\triangle NaS \sim \triangle NS'a'$ from the similarity of triangles inverted through a vertex, so $\angle NS'a'$ is a right angle as well. From this we see that $a'$ lies on the plane perpendicular to $NS'$ which passes through $S'$. Now though, observe that $S'=0$ is the origin, since all the necessary triangles are congruent right isosceles, so in fact $a'$ is on the plane $\mathbb{C}$. As in the construction of $a'$, necessarily $a,a',N$ are all colinear, and since $a$ is on the sphere $\Sigma$ while $a'$ is on the plane $\mathbb{C}$, then $a'$ is exactly the stereographic projection of $a$, so $a'=\hat{a}$.

Since the inversion of the inversion is just the original point, the reverse direction is immediate from the above. Let $\hat{a}$ be any point on $\mathbb{C}$, then let $a$ be the unique point on $\Sigma$ which is projected to $\hat{a}$. By the above, when inverting through $K$, we get $a'=\hat{a}$, but then $\hat{a}'=a''=a$.