I tried to prove in the complex plane that a transformation of a line not passing through the center of a inversion is a circle passing through the center. I just want to know, if what I wrote is right and not false.
We have a line $z(\frac {a-ib}{2})+\bar{z}(\frac{a+ib}{2})+c=0$. To simplify we put: $ w=\frac{a+ib}{2}$ and $\bar{w}=\frac {a-ib}{2}$. So,$d:z\bar{w}+\bar{z}w+c=0$. We know that an image of z is $z'=\frac{1}{\bar{z}} \nonumber$, if the ray of a circle of inversion is 1. So:
$ z\bar{w}+\bar{z}w+c=0 . \text { Dividing by } z\bar{z}: \ \\ \frac {1}{\bar z}\bar{w}+\frac{1}{z}w+\frac {c}{z\bar{z}}=0 \nonumber \\ z'\bar{w}+\bar{z'}w+c\bar{z'}z'=0 \nonumber \\ \frac {z'\bar{w}}{c}+\frac {\bar{z'w}}{c}+\bar{z'}z'=0 \nonumber \\ z'(\frac{\bar{w}}{c})+\bar{z'}(\frac {w}{c})+\bar {z'}{z'}=0 \nonumber $
which is a circle with center $-\frac {\bar{w}}{c}$
In response to a comment to the Q from the OP.
$z\bar w+\bar z w+c=0 \iff 2Re (z\bar w)+c=0.$ So in order that this represents a line, and not the empty set nor a single point, we must have $c\in \Bbb R$ and $w\ne 0$. And in order that $0$ is not on this line we must have $c\ne 0.$
Let $z'=x+iy$ and $w=a+ib$ with $x,y,a,b \in \Bbb R.$ Then because $1/c\in \Bbb R$ we have, from the last displayed line in the Q, $$z'\bar w/c+\bar z' w/c+z'\bar z'=0 \iff$$ $$\iff (1/c)\cdot 2Re (z'\bar w)+|z|^2=0 \iff$$ $$\iff (1/c)\cdot 2(xa+yb)+x^2+y^2=0 \iff$$ $$\iff (x+a/c)^2+(y+b/c)^2=(a^2+b^2)/c^2 \iff$$ $$\iff |z'+w/c|^2=|w|^2/c^2\iff$$ $$\iff |z'+w/c|=|w|/|c|.$$ Since $|w|\ne 0,$ that is the equation of a circle $C$ centered at $-w/c$ (not $-\bar w/c$ as in the Q ) with radius $|w|/|c|.$
Note: The equation for $C$ is satisfied if $z'=0$ but there is no $z$ on the line that inverts to $0.$ The image of the line under the inversion is $C\setminus \{0\}.$