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I've got the following set: $\{|z-a|=k|z-b|\}$, where z is a complex number, a an b are fixed, and $k>0$,$k \ne 1$. I need to prove that this is a circle (called Apollonius circle). I also have to prove that this circle's radius is equal to $k|a-b||1-k^2|^{-1}$ and it's centre is $(a-k^2b)(1-k^2)^{-1}$. I don't know what to do, I've tried to work with analytic circle equation ($|z-c|^2=r^2$), substituting given radius and center, but it didn't work. I also tried to square both sides of the first given equation ($|z-a|=k|z-b|$), which usually works with complex number, but also didn't get any result... Can somebody show m how to solve this problem? I will be very grateful.

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Write it as $(z-a)\overline{(z-a)} = k^2(z-b)\overline{(z-b)}$ and transform to the form $(z-c)\overline{(z-c)}=r^2$. –  njguliyev Oct 9 '13 at 18:12

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The algebra is messy but straightforward. Let $z=x+i y$, $a=a_r+i a_i$ and $b=b_r+i b_i$. Square both sides of the defining equation to get

$$(x-a_r)^2+(y-a_i)^2 = k^2 (x-b_r)^2 + k^2 (y-b_i)^2$$

Rearrange and expand to get

$$(1-k^2) x^2 + (1-k^2) y^2 - 2 (a_r-k^2 b_r) x - 2 (a_i - k^2 b_r) y + |a|^2-k^2 |b|^2 = 0$$

Now complete the square. This is where it gets messy.

$$\begin{align}(1-k^2) \left ( x-\frac{a_r-k^2 b_r}{1-k^2}\right )^2 + (1-k^2) \left ( y-\frac{a_i-k^2 b_i}{1-k^2}\right )^2 \\= \frac{(a_r-k^2 b_r)^2+(a_i-k^2 b_i)^2}{1-k^2} - (|a|^2-k^2 |b|^2)\\ = \frac{|a|^2 + k^4 |b|^2 - 2 k^2 (a_r b_r+a_i b_i) - (|a|^2-k^2 |b|^2)(1-k^2)}{1-k^2}\\ = \frac{k^2 (|a|^2+|b|^2) - 2 k^2 (a_r b_r+a_i b_i)}{1-k^2}\\ = \frac{k^2}{1-k^2} |a-b|^2\end{align}$$

From here, I hope it is clear that the above reduces to, in complex notation:

$$\left |z-\frac{a-k^2 b}{1-k^2}\right| = \frac{k}{1-k^2} |a-b|$$

So, a circle of center $$\frac{a-k^2 b}{1-k^2}$$ and radius $$\frac{k}{1-k^2} |a-b|$$

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Thank you so much for your help and effort! –  Anne Oct 9 '13 at 22:05

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