# determination of a circle in the complex plane

If A$(z_1)$ and B$(z_2)$ are two fixed points in the Argand Plane and P$(z)$ is a variable point satisfying $|z-z_1| = k|z -z_2|$ then determine the locus of P

This is how I I proceeded with the problem :-

$|z-z_1| = k|z-z_2|$

Therefore ,this can be written as

$PA = kPB$

or we can say

$\frac{PA}{PB} = k$

I do not understand what to conclude from here . according to the solution of the problem , the previous equation indicates that the locus of P is a circle (or P lies on a circle) Now how does that make it a circle ? Please help

• The solution is not right. For $k=1$ the locus is the line orthogonal to the segment $AB$ and through its midpoint. – Fimpellizieri Jan 29 '18 at 13:53
• The locus for $k \ne 1$ is a circle of Apollonius, see this answer for the complete derivation. – dxiv Jan 29 '18 at 16:40

## 1 Answer

Actually, this is a circle unless $k=1$ (assuming that $k>0$).

Take $a,b,c,d\in\mathbb R$ such that $z_1=a+bi$ and $z_2=c+di$. Then, for each $x,y\in\mathbb R$,\begin{multline}|(x+yi)-(a+bi)|=k|(x+yi)-(c+di)|\iff\\\iff(x-a)^2+(y-b)^2=k^2\bigl((x-c)^2+(y-d)^2\bigr).\end{multline}Now, expand this equality and you'll get the equation of a circle (except if $k=1$, in which case you'll get a straight line).

• Oh alright I get it now . Thanks for helping !! – Aditi Jan 29 '18 at 14:00