# Circlines. How to solve questions on circles and lines in the complex plane.

I have different equations in my notes but I'm still unclear how to use them in these past few questions. I know things like:

The circle with centre $z_0$ and radius $r$ is the locus of points with equation $$\left|z-z_0\right|=r$$

For distinct points $z_1, z_2 \in \mathbb{C}$ every circle $\left|z-z_0\right|=r$ can be written in the form $$\left|\frac{z-z_1}{z-z_2}\right|=\lambda,\quad\lambda\gt0\quad(\mathtt{whatever\;that\;means...})$$

Here is the question I have that I don't even know where to begin. Find the centre and radius of the following circle $$\left|z+1\right|=4\left|z-1\right|$$

Given points $A,B$ in a plane $P$ and a positive constant $c\ne1$, the locus of points $Z$ in $P$ such that $|ZA|=c|ZB|$ is a circle whose centre lies on the line through $A$ and $B$.
In this case you have $A=(-1,0)$ and $B=(1,0)$ so the line is the $x$-axis. The equation $|ZA|=4|ZB|$ determines two points on the $x$-axis, namely $Z_1=(\frac35,0)$ and $Z_2=(\frac53,0)$. Therefore $Z_1Z_2$ is a diameter of the circle, the centre is $(\frac{17}{15},0)$ and the radius is $\frac8{15}$.
If you don't know the relevant geometry, the fallback method is to substitute $z=x+iy$ and do lots of algebra. In this case you will start with $$|z+1|^2=16|z-1|^2$$ so $$(x+1)^2+y^2=16((x-1)^2+y^2)\ ,$$ and I leave the rest to you.
• Wow, The geometry part was making sense but I'm not sure how you got the numbers for $Z_1$ and $Z_2$. But I followed you're algebra reasoning and did get $z_0$ to be $\left(\frac{17}{15},0\right)$ and $r$ to be $\frac{8}{15}$ So I guess I'll keep to that approach. But what if the 4 was an unknown constant $\lambda$? How would $\vert z+1\vert = \lambda\vert z-1\vert$ be dealt with? to first find $\lambda$ but how? – MRT Feb 16 '17 at 0:18
• You need $x$ between $-1$ and $1$ such that $|x+1|=\lambda|x-1|$, which is $(x+1)=\lambda(1-x)$ and you can solve. The other $x$ value will be greater than $1$ if $\lambda>1$, or less than $-1$ if $0<\lambda<1$. In either case the absolute values will work out a bit differently. – David Feb 16 '17 at 0:52