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I am having trouble with this question:

Suppose that $k|z-z_{1}|=l|z-z_{2}|$, where $k\neq l$ and $k,l>0$, $k, l\in\mathbb{R}$. Show that the locus of z in the Argand diagram is a circle with the centre $\frac{k^{2}z_{1}-l^{2}z_{2}}{k^{2}-l^{2}}$ and radius $\frac{kl|z_{2}-z_{1}|}{|k^{2}-l^{2}|}$ by letting $z=x+iy$

I used the identity that a circle with diametrical endpoints $(x_{a}, y_{a})$ and $(x_{b}, y_{b})$ can be written as $$(x-x_{a})(x-x_{b})+(y-y_{a})(y-y_{b})=0$$

Letting $z_{1}=a+ib$ and $z_{2}=u+iv$, the identity gave me $$\left(x-\frac{ak+lu}{k+l}\right)\left(x-\frac{ak-lu}{k-l}\right)+\left(y-\frac{kb+lv}{k+l}\right)\left(y-\frac{kb-lv}{k-l}\right)=0$$

However, I am unsure how to distinguish $x_{a}$ and $x_{b}$, $y_{a}$ and $y_{b}$ (from the identity) in this case i.e I am not sure how to match the real and imaginary parts of the diametrical endpoints. I guessed correctly that the real and imaginary parts had the same denominators, but why is this the case? This method also involved extremely heavy algebra, is there a way to do it that is shorter and doesn't use the identity?

Thanks, BaroqueFreak

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    The locus of points with constant ratio between distances to two fixed points is a circle, whose center and radius are fairly easy to derive geometrically. Lookup the [circles of Apollonius](https://en.wikipedia.org/wiki/Apollonian_circles). – 2017-02-10
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    Yes, but the question specifically asks by letting $z=x+iy$ :( – 2017-02-10
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    I read that more as a hint than a requirement. What you are asked to prove is essentially $\left| z - \frac{k^{2}z_{1}-l^{2}z_{2}}{k^{2}-l^{2}} \right| = \frac{kl|z_{2}-z_{1}|}{|k^{2}-l^{2}|}\,$. For a geometrically-less approach, you could try and adapt [this](http://math.stackexchange.com/a/2091243/291201). – 2017-02-10

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