Hello, prasum!
$\displaystyle \text{The centre of circle represented by: }\;z+1 \:=\:2z1$
$\displaystyle z+1\:=\:z  (\text{}1)$ is the distance from $\displaystyle \,z$ to (1,0).
$\displaystyle z1$ is the distance from $\displaystyle \,z$ to (1,0).
Code:
 P
 o
 * *
*  *
A *  * B
  o     +     o  
1  1

We want points $\displaystyle P(x,y)$ so that: .$\displaystyle \overline{PA} \;=\;2\!\cdot\!\overline{PB}$
[This is known as the Circle of Apollonius.]
We have: .$\displaystyle \sqrt{(x+1)^2 + y^2} \;=\;2\sqrt{(x1)^2+y^2}$
Square: .$\displaystyle (x+ 1)^2 + y^2 \;=\;4\left[(x1)^2 + y^2\right]$
which simplifies to: .$\displaystyle 3x^2  10x + 3y^2 \;=\;3$
. . . . . . . . . . . . . . . . $\displaystyle x^2  \frac{10}{3}x + y^2 \;=\;1$
Complete the square: .$\displaystyle x^2  \frac{10}{3}x + \left(\frac{5}{3}\right)^2 + y^2 \;=\;1 + \left(\frac{5}{3}\right)^2 $
. . . . . . . . . . . . . . . . . . . . . $\displaystyle \left(x  \frac{5}{3}\right)^2 + y^2 \;=\;\frac{16}{9}$
Therefore, the circle has center $\displaystyle (\frac{5}{3},\:0)$ radius $\displaystyle \frac{4}{3}$