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In the book I'm reading, one of the exercises starts with mentioning that $\left|\frac{z-z_1}{z-z_2}\right|=c$, where $c\neq 1$ is a constant, is an equation of the circle.

It seems clear to me that $|z-z_1|=c$ is an equation of the circle in a complex plane but I can't figure out why the previous formula would also be an equation of the circle. Help with understanding would be appreciated.

Dave
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Storyteller011
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    Are you familiar with Möbius transformations? – njguliyev Sep 27 '13 at 19:21
  • No. These exercises I am going through are at the fairly introductoryu level of introducing complex numbers. I am familiar with polar and exponential representations, basic identities of hyperbolic equations etc but not with projective transformations. – Storyteller011 Sep 27 '13 at 19:26

3 Answers3

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A brute force approach : Write $z = x+iy, z_1 = x_1+iy_1, z_2 = x_2+iy_2$, and simplify the formula $$ |z-z_1|^2 = c^2|z-z_2|^2 $$ You will end up with an equation of the form $$ (1-c^2)x^2 + (1-c^2)y^2 + 2\alpha x + 2\beta y +\gamma = 0 $$ for some constants $\alpha, \beta, \gamma$. Since $c\neq 1$, you can divide by $(1-c^2)$ (Note that $c$ is a real number and $\neq -1$ because of what you started with). This will give you the equation of a circle.

Prahlad Vaidyanathan
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It is the same as $|z-z_1|^2=c^2|z-z_2|^2$.
Write $z=x+iy$, $z_1=x_1+iy_1$ and so on.
Expand the squared absolute values, and simplify.

Empy2
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From an understanding point of view, if $ |z−z_1|=c $ is a circle. Then $ {∣z−z_1|\over|z−z_2∣}=c$, where c≠1 can be written as $|z−z_1|=c|z−z_2|$ which clearly looks like $ |z−z_1|$ is a circle scaled by a factor of c from another circle $ |z−z_2|=k$.

Hope that makes it clearer.

MANI
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Amogh Jahagirdar
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