What is the radius of the circle $|z - z_0| = \rho |z - z_1|$? A conjecture based on street fighting mathematics.

Question

What is the radius of the circle $|z - z_0| = \rho |z - z_1|$? My conjecture is $$\text{radius} = |z_1 - z_0| \cdot \frac {\rho ^2}{(\rho -1)^2(\rho + 1)^2}$$ which I request help verifying, proving, or correcting.

Note: Formulas exist; I'm asking for help with this approach, as described below.

Let $d = |z_1 - z_0|$. Using the approach taught in Street Fighting Mathematics, I deduced the following:

1. The radius is obviously a function of $d$ and $\rho$.
2. It must be undefined when $\rho = 1$.
3. This also implies that $\rho$ is unitless.
4. It must be linear in $d$, since the radius is in units of linear distance, and $d$ is as well (whereas $\rho$ is unit less).
5. It must be symmetric when switching $z_0, z_1$, implying $f(\rho) = f(\frac 1 \rho)$.
6. It must approach $0$ when $\rho \to 0$ or $\rho \to \infty$.
7. It must approach $\infty$ when $\rho$ approaches $1$ (illustrated by Circle of Appolonius diagrams).

Asking What function $f(x) = f(1/x)$ has limits $\lim_{x \to 0} f(x) = 0, \lim_{x \to \infty} f(x) = 0$, and $\lim_{x \to 1} f(x) = \infty$? , I got some candidate functions, took the simplest, realized I needed to square $\rho$ to makes things work (which is intuitive given that distance requires taking a square root, and so e.g. the standard equation of circles is in $r^2$), and got this candidate: $$\text{radius} = \frac d {\rho^2 + \frac 1 {\rho^2} - 2}\\ = d \cdot \frac {\rho ^2}{(\rho -1)^2(\rho + 1)^2}.$$

Can you verify or correct this? Is there a way I could develop this approach into a proof?

Answer 1

You have made an amazing list of 7 observations on the radius of the circle $|z - z_0| = \rho |z - z_1|$.

Unfortunately, they are not enough to single out the right formula for the radius.

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Here is another observation.

Fix $d=|z_0- z_1|$. When $\rho$ is very small, $z$ is very near to $z_0$. Hence, $|z-z_1|$ is roughly the same as $|z_0-z_1|=d$. So, $|z-z_0|=\rho|z-z_1|\approx\rho d$. That is, the radius $r\approx\rho d$. More precisely, when $\rho\to0$, $\frac{r}{\rho d}\to1$.

Since $r$ is linear in $d$, let $r(\rho,d)=g(\rho)d$ for some function $g(\rho)$, $\rho>0$. Then our observations mean

• $\lim_{\rho \to 0} \frac{g(\rho)}\rho = 1$,
• $\lim_{\rho \to \infty} g(\rho) = 0$,
• $\lim_{\rho \to 1} g(\rho) =\infty$,
• $g(\rho)$ is continuous except at $\rho=1$, where it is probably undefined.
• $g(\rho) = g(1/\rho).$

The simplest function that satisfies the observations above is $$g(\rho)=\frac1{|\rho-\frac1\rho|}.$$

Hence, we could hope the answer might be

$$r(\rho,d)=\frac d{|\rho-\frac1\rho|}.$$

Fortunately and magically, it is correct!

Answer 2

I won't talk much about the "street fighting mathematics" - I appreciate the lack of rigour could on one hand make you go down the slippery slope but could in other cases provide valuable insight. Which is to say your mileage may vary. However, I wanted to give a different sort of intuition here, which I think is still in the spirit of the "street fighting mathematics".

Note that the circle that has the property $|z-z_0|=\rho|z-z_1|$ has two points of intersection with the line $(z_0,z_1)$ and both of those points have to satisfy the same equation. So, forget all about circles for now! Call those points $z_i$ ("inner") and $z_o$ ("outer"). Let's say $\rho<1$ so $z_o$ is on the "right" of $z_1$ with respect to $z_0$.

Now: $|z_i-z_0|=\rho|z_i-z_1|$ so $z_i$ divides the segment $[z_0,z_1]$ in the ratio $1:\rho$. If $|z_i-z_0|=x$ then $|z_i-z_1|=d-x$ and so you need to solve the (linear) equation $d-x=\rho x$ which has a solution $x=\frac{d}{1+\rho}$. Similarly, for $z_o$, you would call $|z_0-z_o|=y$, which would imply $y-d=\rho y$ and would have a solution $y=\frac{d}{1-\rho}$.

Okay. Back to circles: the segment $[z_i,z_o]$ is the diameter of the circle, and so your radius is just $\frac{y-x}{2}$, which is:

$$\frac{1}{2}\left(\frac{d}{1-\rho}-\frac{d}{1+\rho}\right)=\frac{\rho d}{1-\rho^2}$$

Now, when $\rho>1$, you can repeat the entire process ... or you can use your observation that you can substitute $\rho$ with $\frac{1}{\rho}$, so the radius is:

$$\frac{\frac{1}{\rho}d}{1-\left(\frac{1}{\rho}\right)^2}=\frac{\rho d}{\rho^2-1}$$

and that is how we end up with the formula $\frac{\rho d}{|1-\rho^2|}$.