# why does multiplicatively weighted voronoi diagram (mwvd) with 2 sites create a circle?

I want to understand the structure of a multiplicatively weighted voronoi diagram. I found that the bisector between 2 sites is circle shaped, but couldn't formally see it. can someone explain?

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For the benefit of the readers I give a short description: You have a set of cities $p_i$ in the euclidean plane $E:={\mathbb R}^2$, and each of these cities has a given weight $w_i>0$. The "felt distance" $d(z,p_i)$ from an arbitrary point $z\in E$ to the city $p_i$ is defined to be $$d(z,p_i)\ :=\ {|z-p_i| \over w_i}\ ,$$ where $|z-z'|$ denotes the euclidean distance. This means that the "felt distance" to a city with large weight is comparatively small, and conversely, that the $d$-unit-disk of a city with large weight has a large euclidean radius.
Cosider now two cities $p$ and $p'$ with respective weights $w$ and $w'$. The $d$-Voronoi-boundary $\partial$ between these two cities consists of the points $z\in E$ which have the same $d$-distance to $p$ and to $p'$, i.e., it is defined by the equation $${|z-p|\over |z-p'|} \ =\ {w\over w'}\ .\qquad (*)$$ This says that $\partial$ is the locus of all points $z\in E$ for which the ratio $|z-p|/ |z-p'|$ has the a priori given value $\lambda:=w/w'$. It is a theorem of elementary geometry that such a locus is a circle, called the ${\it Apollonian\ circle}$ for the given data $p$, $p'$ and $\lambda$. The simplest proof is by choosing $p=(0,0)$, $p'=(c,0)$. Then the condition $(*)$ becomes $x^2+y^2=\lambda^2\bigl((x-c)^2 + y^2 \bigr)$, which can be simplified to the equation of a circle (or a line).