Locus of intersecting circles I would like to find the locus of $z\in\mathbb{C}$ such that
$$\left| z-a\right|^2=k\left| z-b\right|^2 ,$$
where $k>0$, and $a,b\in\mathbb{C}$.
I found, if $k\neq 1$, the circle
$$\left| z-\dfrac{1}{1-k}(a-kb)\right|^2=\dfrac{k}{(1-k)^2}\left| a-b\right|^2 .$$
If $k=1$, the equidistant line from $a$ and $b$.
My problem is that I had to solve it in $\mathbb{R}^2$ with $z=(x,y)$, ... And I'm almost sure that there is a way to see it more geometrically (that recall the complex inversion?). I tried to consider the map
$$z\longmapsto \dfrac{\left| z-a\right|^2}{\left| z-b\right|^2} $$
But nothing came out.
Do you have any idea please?

You essentially have
$$\lvert z-a\rvert = \sqrt k\,\lvert z-b\rvert$$
which is the Apollonian definition of a circle as the set of all points having constant ratio of distances to two given points.
There are various ways of describing said locus, and I'd say this very formula (or your squared version) is already a definition of the Locus. So if you need a different formula, you have to be more specific about what kind of representation you desire. Should it be center and radius (which will fail for the degenerate case of $k=1$)? Or should it be thee points on the circle defining said circle? Or just some geometric intuition?
