What is the locus of $u|z-P| = v|z-Q|$ where $u \ge 0, v \ge 0, u+v > 0$?
This is a generalization of Identify the plane defined by $|z-2i| = 2|z+3|$.
My result, after some annoyingly complex algebra is this:
Let $P=a+bi$ and $Q = c+di$.
If $u=v$ the locus is the line $$0 =-2x(a-c) -2y(b-d) +(a^2+b^2-c^2-d^2) $$ which is the perpendicular bisector of the line joining $P$ and $Q$.
If $u \ne v$ the locus is the circle with center $\left(\dfrac{au^2-cv^2}{u^2-v^2}, \dfrac{bu^2-dv^2}{u^2-v^2}\right) $ and radius $\dfrac{uv}{u^2-v^2}\sqrt{(a-c)^2+(b-d)^2} $.
My questions are is this correct (especially the second part) and is there a simpler way (probably using vectors) to derive these?
I wouldn't be surprised if this is already here, somewhere.