Image of a locus via stereographic projections

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Yesterday evening I was playing around in my head with stereographic projections and I've come up with this idea. Let $\gamma(t)=(x(t),y(t))$ be a certain curve on a plane. Define a new curve \begin{equation} \hat{\gamma}(t):=\left(\frac{x(t)}{\|\gamma(t)\|^2},\frac{y(t)}{\|\gamma(t)\|^2}\right) \end{equation} The idea behind this is the following. Let $\pi_1$ be the plane in which $\gamma$ lies. Suppose you have a sphere of radius $1/2$ tangent at the origin $A$ of $\pi_1$. The sphere has an antipodal point $B$ with respect to $A$. We can consider two maps: the inverse stereographic projection $\Phi_B^{-1}$ from $\pi_1$ to the sphere with respect to $B$ and the stereographic projection $\Phi_A$ from the sphere to the plane $\pi_2$ tangent to the sphere in $B$ (with respect to the point $A$). Then $\hat{\gamma}$ is defined as \begin{equation} \hat{\gamma}:=\Phi_A\circ\Phi_B^{-1}\circ\gamma \end{equation} If $\gamma(t)=(t,f(t))$ for some function $f$, we get \begin{equation} \hat{\gamma}(t)=\left(\frac{t}{t^2+f(t)^2},\frac{f(t)}{t^2+f(t)^2}\right) \end{equation}

here's the idea with $f(x)=\frac{1}{x}$: enter image description here

and with a parabola $f(x)=x^2$: enter image description here

My question is: Is this a known and used transformation, i.e are there interesting relations between $\gamma$ and $\hat{\gamma}$?

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You are describing inversion in the unit circle:

$$(x,y)\mapsto\frac{1}{x^2+y^2}(x,y)$$

In a polar coordinate system, it preserves direction and inverts radius.

And yes, this concept is used, as the tag here on Math SE indicates. There are of course questions where this tag wasn't added, but an answer still made use of circle inversions in order to simplify the problem. I seem to recall a beautiful post somewhere here on Math SE where the OP asked about some locus related to some iterated packing of circular disks, and the answer transformed the problem using inversions in such a way that this locus became a straight line. Can't find it just now, though.