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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.

marty cohen
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    See here for the complete derivation, with a slightly different notation (the $k$ there is equivalent to your $v/u,$). The locus is indeed either the perpendicular bisector of the segment $PQ,$ for $u=v,$, or a circle of Apollonius having as diameter the two points which divide the segment $PQ$ in ratio $v : u \ne 1$ internally and externally. – dxiv May 15 '17 at 21:58
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    Thanks. I thought of doing it like that answer, but I wanted to have my question be symmetric in the two sides. – marty cohen May 15 '17 at 22:01

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