Centre and radius of the circle $|z-a|/|z-b|=c$

Question

I am trying to calculate the centre and the radius of the circle defined by

$$|z-a|/|z-b|=c$$

for $a$, $b \in \mathbb{C}$, $c \neq 1$. (This is indeed a circle, see this question.)

Answer 1

Let $u_{a,b}$ be the center and $r_{a,b}$ be the radius. Note the question makes no sense if $a=b.$

First show that $u_{a,b}=u_{a-b,0}+b$ and $r_{a,b}=r_{a-b,0}.$

Now, it is not hard to prove that if $z$ is on the circle with $b=0,$ then $z/a$ is on the circle with $a'=1, b'=0.$

So $u_{a,0}=au_{1,0}, r_{a,0}=|a|r_{1,0}.$

So you need to compute $r_{1,0},u_{1,0}$ and you get:

$$r_{a,b}=r_{a-b,0}=|a-b|r_{1,0},\\u_{a,b}=u_{a-b,0}+b=(a-b)u_{1,0}+b$$

The last step is to compute $r_{1,0},u_{1,0}.$ This is a relatively simple problem.

The solutions to: $$\left|\frac{z-1}{z}\right|=c$$ are symmetric on the real line, so the real line is a diameter, and the points where it intersects the real line are $$z=\frac 1{1\pm c},$$ and thus the center and radius are: $$u_{1,0}=\frac12\left(\frac1{1+c}+\frac1{1-c}\right)=\frac1{1-c^2}\\ r_{1,0}=\left|\frac{c}{1-c^2}\right|$$

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The initial argument can be written briefly as:

$$\frac{z-a}{z-b}=\frac{w-1}{w}$$ when $w=\frac{z-b}{a-b},$ or the map is $\phi:w\mapsto (a-b)w +b=z.$ But $\phi$ just scales, translates, and rotates. So it sends the center of a circle to the center of the image circle.

Answer 2

This can be interpreted as the set of points that are $c$ times further away from $a$ than $b$, so we know that the points

$$z_1 = \frac{a+cb}{1+c} \text{ and }z_2 = \frac{a-cb}{1-c} $$

lie on the circle.

These points both lie on the line passing through $a$ and $b$, which by symmetry must contain the centre, so they are diametrically opposite each other. Thus, the centre of the circle is at their average

$$z_*:=\frac{1}{2}(z_1+z_2) = \frac{a-c^2b}{1-c^2}.$$

The radius of the circle is then the distance between $z_*$ and either of these points

$$|z_*-z_1|=|z_*-z_2|=\frac{c|a-b|}{1-c^2}.$$