The claim is that: if a line L does not pass through the centre $q$ of $K$, then inversion in $K$ maps $L$ to a circle that passes through $q$; where $b$ is an arbitrary point on $L$ while $a$ is the intersection of $L$ with the perpendicular line through $q$. But how can I prove this synthetic-geometrically, i.e. without algebra?
Now, I've been thinking about this for a while. By Thales' theorem, drawing a circle with diameter $\tilde{a}q$ will necessarily allow for the constitution of a right trangle $q\tilde{b}\tilde{a}$, and the author mentions that the triangle $q\tilde{b}\tilde{a}$ is similar to the triangle $qab$, and this also follows from Thales' theorem. Furthermore, since $\tilde{b}\tilde{a}$ is a line that intersects perpendicularly a line through $q$, $\tilde{b}$ is necessarily the circle inversion of some point, which happens to be $b$ if we draw the two tangent lines of $b$ with $K$, but this seems to happen completely coincidentally.
The proof is easier with algebra, but how come $\tilde{b}$ necessarily produces the inversion of $b$ in $K$? Any ideas of synthetic geometrical theorems which could help?
If this helps anyone, here are the geometrical constructions following the fact that both $\tilde{b},\tilde{a}$ are inversions:
Figure [3], by the way, is from Needham's Visual Complex Analysis, page 127.
EDIT: thanks to user Oussema for pointing out the solution to similarity between $qba$ and $q\tilde{b}\tilde{a}$!
Best Answer
Thanks to user Oussema for an important idea here. Create the following construction:
Where $C$ is a circle with center $c$, and $c$ is a midpoint on $bD$. Then, by Thales' theorem, any intersections $I1, I2$ of the circle C with circle $K$ will generate tangents of $K$ with $b$ (since the angles are, by Thales theorem, right angled), and therefore, the intersection of $I1I2$ with $bD$ will by definition be the inversion of $b$ in $K$, $b^*$!