Let $u, v \in \mathbb{C}$. Consider the circles centered at $u, v$ such that they intersect exactly at the point $\frac{u+v}{2}$, i.e. $C_u = \partial D_r(u), C_v = \partial D_r(v)$ where $D_r$ denotes the disk of radius $r$ and $r = |\frac{u+v}{2} - u| = |\frac{u+v}{2} - v|$. These circles are given by the equations $|z - u|^2 = r^2$ and $|z - v|^2 = r^2$ which are the same as
$$ z\bar{z} - u\bar{z} - \bar{u}z + u\bar{u} - r^2 = 0, \\z\bar{z} - v\bar{z} - \bar{v}z + v\bar{v} - r^2 = 0.$$
I now want to prove the following fact: The equation of the line $|z - u| = |z - v|$, i.e. $$-(v - u)\overline{z} + -\overline{(v-u)} z + (|v|^2 - |u|^2) = 0$$ arises as the limit of these equations as $|u|, |v|, |u - v| \rightarrow \infty$.
I truly have no idea how to even begin making sense of this. Geometrically, this seems intuitive enough, a circle of infinite radius should be a line. Substituting $r$ into the first equation, we get
$$ z\bar{z} - u\bar{z} - \bar{u}z + u\bar{u} - \frac{(v-u)\bar{v} + (u-v)\bar{u}}{4} = 0,$$
but I don't understand how I can get $v, \bar{v}$ to be a factor in front of some $z$ term (as in the equation I am trying to get) just by taking a limit. What I find particularly troubling is that I get the desired equation simply by subtracting both of the circle equations from each other, but that doesn't feel like I'm doing any sort of limiting process.