Writing circles as $|z-a| = \lambda |z-b|$ for the same $a,b$

Question

My problem is in the context of the complex plane. I want to know if given two disjoint, not concentric circles $C_1,C_2\subset \mathbb{C}$, can you find $a,b\in \mathbb{C}$ such that $$C_1=\{z\in \mathbb{C} : |z-a| = \lambda_1|z-b|\}$$ $$C_2 = \{z\in \mathbb{C}:|z-a| =\lambda_2|z-b|\}$$ If the problem is true, how can you find them or how can you prove the existence. If it's false, is there a counterexample?

Answer 1

The answer is yes.

Without loss of generality, we may suppose that $$C_1=\{z\in \mathbb{C} : |z| = 1\},\quad C_2 = \{z\in \mathbb{C}:|z-p| =r\}$$ where $p\gt 0\in\mathbb R$ and $r\ge 1\in\mathbb R$ such that $|p-r|\gt 1$.

By the way, noting that we have to have $\lambda_i\not=1$ and $\lambda_i\gt 0$, we have $$\begin{align}&|z-a|=\lambda|z-b|\\&\iff (z-a)(\bar z-\bar a)=\lambda^2 (z-b)(\bar z-\bar b)\\&\iff z\bar z-\bar az-a\bar z+a\bar a=\lambda^2(z\bar z-\bar bz-b\bar z+b\bar b)\\&\iff (1-\lambda^2)z\bar z+(\lambda^2 \bar b-\bar a)z+(b\lambda^2-a)\bar z+a\bar a-\lambda^2b\bar b=0\\&\iff z\bar z+\frac{\lambda^2\bar b-\bar a}{1-\lambda^2}z+\frac{b\lambda^2-a}{1-\lambda^2}\bar z+\frac{a\bar a-\lambda^2b\bar b}{1-\lambda^2}=0\\&\iff \left(z+\frac{b\lambda^2-a}{1-\lambda^2}\right)\left(\bar z+\frac{\lambda^2\bar b-\bar a}{1-\lambda^2}\right)-\frac{(b\lambda^2-a)(\lambda^2\bar b-\bar a)}{(1-\lambda^2)^2}+\frac{a\bar a-\lambda^2b\bar b}{1-\lambda^2}=0\\&\iff \left(z+\frac{b\lambda^2-a}{1-\lambda^2}\right)\left(\bar z+\frac{\lambda^2\bar b-\bar a}{1-\lambda^2}\right)=\frac{(a-b)(\bar a-\bar b)\lambda^2}{(1-\lambda^2)^2}\\&\iff \left|z-\frac{a-b\lambda^2}{1-\lambda^2}\right|=\frac{|a-b|\lambda}{|1-\lambda^2|}\end{align}$$

So, you want to know if we can find $a,b\in\mathbb C$ and $\lambda_1,\lambda_2\in\mathbb R$ such that $$\frac{a-b\lambda_1^2}{1-\lambda_1^2}=0,\quad \frac{|a-b|\lambda_1}{|1-\lambda_1^2|}=1,\quad \frac{a-b\lambda_2^2}{1-\lambda_2^2}=p,\quad \frac{|a-b|\lambda_2}{|1-\lambda_2^2|}=r\tag1$$ $$\lambda_i\gt 0,\quad \lambda_i\not=1,\quad \lambda_1\not=\lambda_2,\quad a\not=b\tag2$$

when we are given $p\gt 0\in\mathbb R$ and $r\ge 1\in\mathbb R$ such that $|p-r|\gt 1$.

Now $$\begin{align}(1)&\Rightarrow \lambda_1^2=\frac ab,\quad |a-b|\lambda_1=\left|1-\frac ab\right|,\quad a-b\lambda_2^2=p(1-\lambda_2^2),\quad |1-\lambda_1^2|\lambda_2=r\lambda_1|1-\lambda_2^2|\\&\Rightarrow a=\frac{1}{\bar b},\quad \lambda_1=\frac{1}{|b|},\quad \lambda_2^2=\frac{p-\frac{1}{\bar b}}{p-b},\quad \left|1-\frac{1}{|b|^2}\right|^2\cdot \frac{p-\frac{1}{\bar b}}{p-b}=r^2\frac{1}{|b|^2}\left|1-\frac{p-\frac{1}{\bar b}}{p-b}\right|^2\\&\Rightarrow \frac{||b|^2-1|^2}{|b|^2}\cdot \frac{p-\frac{1}{\bar b}}{p-b}=\frac{r^2|1-|b|^2|^2}{|\bar b|^2(p-b)(p-\bar b)}\\&\Rightarrow |b|=1\quad\text{or}\quad \left(p-\frac{1}{\bar b}\right)(p-\bar b)=r^2\\&\Rightarrow |b|=1\quad\text{or}\quad p^2-(x-yi)p-\frac{p(x+yi)}{x^2+y^2}+1=r^2\quad\text{where $\ b=x+yi\ (x,y\in\mathbb R)$}\\&\Rightarrow |b|=1\quad\text{or}\quad pb^2+(r^2-1-p^2)b+p=0\\&\Rightarrow |b|=1\quad\text{or}\quad b=\frac{p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}}{2p}\end{align}$$

Here, $|b|\not=1$ because $$|b|=1\quad \Rightarrow \quad\lambda_1=1\quad\text{and}\quad \lambda_2^2=\frac{p-b}{p-b}=1\quad\Rightarrow\quad \lambda_1=\lambda_2$$

Also, we have

$$\begin{align}p-b\le 0&\Rightarrow p-\frac{p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}}{2p}\le 0\\&\Rightarrow 2p^2-\left(p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}\right)\le 0\\&\Rightarrow p^2-1+r^2\le \pm\sqrt{(r^2-p^2-1)^2-4p^2}\\&\Rightarrow (p^2-1+r^2)^2\le (r^2-p^2-1)^2-4p^2\\&\Rightarrow r^2\le 0\end{align}$$

So, since we have $p-b\gt 0$, we have $p-\frac 1b\gt 0$ because $b_{\pm}=\frac{1}{b_{\mp}}$.

Thus, since $\frac{p-\frac 1b}{p-b}\gt 0$, we have $$(1)\Rightarrow a=\frac{1}{b},\quad \lambda_1=\frac{1}{|b|},\quad \lambda_2=\sqrt{\frac{p-\frac{1}{b}}{p-b}}$$ where $$b=\frac{p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}}{2p}.$$

Note here that $b\not=0$ and $(r^2-p^2-1)^2-4p^2\gt 0$ because

$$b=0\Rightarrow p=0$$ and $$\begin{align}(r^2-p^2-1)^2-4p^2&=(r^2-p^2-1)-(2p)^2\\&=(r^2-p^2-1+2p)(r^2-p^2-1-2p)\\&=(r^2-(p-1)^2)(r^2-(p+1)^2)\\&=(r+p-1)(r-p+1)(r+p+1)(r-p-1)\\&=(r+p-1)(r+p+1)((p-r)^2-1)\\&\gt 0\end{align}$$

Now let us see if these satisfy $(2)$.

First, $\lambda_i\gt 0$ because $$\lambda_1=\frac{1}{|b|}\gt 0,\quad \lambda_2=\sqrt{\frac{p-\frac 1b}{p-b}}\gt 0.$$

Second, $\lambda_1\not=\lambda_2$ because $$\begin{align}\lambda_1=\lambda_2&\Rightarrow \lambda_1^2=\lambda_2^2\\&\Rightarrow \frac{1}{b^2}=\frac{p-\frac 1b}{p-b}\\&\Rightarrow p-b=pb^2-b\\&\Rightarrow |b|=1\\&\Rightarrow \frac{p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}}{2p}=\pm 1\\&\Rightarrow \pm\sqrt{(r^2-p^2-1)^2-4p^2}=\pm 2p-(p^2+1-r^2)\\&\Rightarrow (r^2-p^2-1)^2-4p^2=4p^2+(p^2+1-r^2)^2\mp 4p(p^2+1-r^2)\\&\Rightarrow -2p=\mp (p^2+1-r^2)\\&\Rightarrow r^2=(p-1)^2,(p+1)^2\\&\Rightarrow r=|p-1|, p+1\\&\Rightarrow |p-|p-1||\gt 1\quad\text{or}\quad |p-(p+1)|\gt 1\\&\Rightarrow p\lt 0\quad \text{or}\quad 1\gt 1\end{align}$$

Third, $\lambda_i\not=1$ and $a\not=b$ because
$$\lambda_1=1\quad\text{or}\quad \lambda_2=1\quad\text{or}\quad a=b\quad\Rightarrow\quad |b|=1.$$

As a result, $$(1)(2)\iff (a,b,\lambda_1,\lambda_2)=\left(b_+,b_-,\frac{1}{|b_-|},\sqrt{\frac{p-b_+}{p-b_-}}\right)\quad\text{or}\quad \left(b_-,b_+,\frac{1}{|b_+|},\sqrt{\frac{p-b_-}{p-b_+}}\right)$$ where $$b_{\pm}=\frac{p^2+1-r^2\pm\sqrt{(r^2-p^2-1)^2-4p^2}}{2p}.$$