This is a proof that an image of a circle not passing through the center of inversion is a circle not passing through the center. I want to know if my proof is correct and coherent.
Let be a circle of inversion $C(O,1)$. Let be another circle $C_1(O,r)$ with $O$ of affix $w$ of equation $z\bar{z}-z\bar{w}-\bar{z}w=r^2-\mid w \mid$, where $r^2 \neq \mid w \mid$. Let be $w=a+ib$ and $z=x+iy$ with $x,y,a,b \in \mathbb {R}$. As we know, inverse of z is $z'= \frac{1}{\bar{z}}$. So we can determine now an equation of the image of $C_1$:
$ z\bar{z}-z\bar{w}-\bar{z}w=r^2-\mid w \mid \nonumber \\ 1-2\Re(\bar{w}z')=\mid z' \mid^2(r^2-\mid w \mid). \nonumber $
Let $z'=x'+iy'$ with $x',y' \in \mathbb {R}$:
$ \dfrac{1-2ax'-2by'}{(r^2-\mid w \mid)}= \dfrac{(x'^2+y'^2)(r^2-\mid w \mid)}{(r^2-\mid w \mid)}\nonumber$ $\iff x'^2+\dfrac {2ax'}{r^2-\mid w \mid}+y'^2+\dfrac {2by'}{r^2-\mid w \mid}=\dfrac {1}{r^2-\mid w \mid} \nonumber \\ \iff (x'+\dfrac {a}{r^2-\mid w \mid})^2-\dfrac{a^2}{(r^2-\mid w \mid)^2}+(y'+\dfrac{b}{r^2-\mid w \mid})^2-\dfrac{b^2}{(r^2-\mid w \mid)^2}=\dfrac {1}{r^2-\mid w \mid} \nonumber \\ \iff \left| z'+\dfrac {w}{r^2-\mid w \mid}\right|=\sqrt{\dfrac {1}{r^2-\mid w \mid}+\dfrac{\mid w \mid^2}{(r^2-\mid w \mid)^2}} $
Which is an equation of a circle not passing through $O(0,0)$ centered in $-\dfrac{w}{r^2-\mid w \mid}$ with radius $\sqrt{\dfrac {1}{r^2-\mid w \mid}+\dfrac{\mid w \mid^2}{(r^2-\mid w \mid)^2}}$ if $w \neq 0$.
The approach is correct, as you first note that $z'=\frac{1}{\bar z}\iff z={\frac{1}{\bar z'}}$, $q=(z')^{-1}\in \mathbb{C}$ and substituting this into the equation gives, as you have shown;
$${\frac{1}{z'\bar z'}}-(\frac{\bar{w}}{\bar z'}+\frac{w}{z'})=r^2-| w |$$ $$| (z')^{-1}|^2 -\overline{({z}')^{-1}{w}}-({z}')^{-1}{w}-r^2+| w |=0$$ $$| q|^2 -\overline{qw}-qw-r^2+| w |=0 \quad(1)$$
which in fact, already represents a circle iff $|w|^2>|w|-r^2$. The equation above $(1)$ is actually called a Moebius circle. Your proof shows the same, also determining the center and radius, without need of the Moebius concept. But the condition for $|w|^2$ must apply: as you can see, the radius
$R=\sqrt{\dfrac {1}{r^2-| w |}+\dfrac{| w |^2}{(r^2-| w |)^2}}=\sqrt{\dfrac{r^2-|w|+| w |^2}{(r^2-| w |)^2}}$ is undefined, if $r^2+|w|^2 \ngtr |w|$.
A late answer, hope it is still useful. Look for sources about Moebius circles or Moebius transforms to find additional information, if wanted.