Question

The centre of the circle represented by $\left| \mathbf{z}\text{ }+\text{ }\mathbf{1} \right|\text{ }=\text{ }\mathbf{2}\left| \mathbf{z}\text{ }\text{ }\mathbf{1} \right|$in the complex plane is:
A. $\left( 0,0 \right)$
B. $\left( \dfrac{5}{3},0 \right)$
C. $\left( \dfrac{1}{3},0 \right)$
D. None of these

Answer 1

A circle is a focus of such a point which maintains a constant distance from a fixed given point. Let the fixed point be A(a,b), let the point whose focus is to be found be P(x,y) and it maintains a distance r from the given point is nothing but the modulus of the complex number. So, in complex coordinate system, the circle can be represented as
$\left| z \right|\text{ }=\text{ }r,$
Now we shift the origin back to its original position. So, know the equation becomes:
$\left| z-a \right|\text{ }=\text{ }r$
Where $r$ represents the radius of the circle.

Complete step by step answer:
Step1: We have given that:
$\left| z+1 \right|\text{ }=\text{ }2\left| z-1 \right|\ldots \ldots \ldots \ldots .\text{ }\left( A \right)$
in a complex plane
Let,
$z=\text{ }x+iy$
Here x represents the real part and y represents the imaginary part.
while x and y are both real numbers.

Step2: Put the value of z in equation (A):
$\left| x+iy+1 \right|\text{ }=\text{ }2\left| \left( x+iy \right)\text{ }\text{ }1 \right|$
rearranging real and imaginary parts:
|(x+1) + 1y| = 2|(x-1) + 1y|
$\left| \left( x+1 \right)\text{ }+\text{ }1y \right|=2\left| \left( x-1 \right)+\text{ }1y \right|$
solving the magnitude, implies:
$\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-1)}^{2}}+{{y}^{2}}}$
Taking square on both hand sides, we get
${{(x+1)}^{2}}+{{y}^{2}}=4\left( {{(x-1)}^{2}}+{{y}^{2}} \right)$

Let us now use the following identities:
${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
After using identities, we get:
${{x}^{2}}+1+2x+{{y}^{2}}=4({{x}^{2}}+1-2x+{{y}^{2}})$
which implies that
${{x}^{2}}+1+2x+{{y}^{2}}=(4{{x}^{2}}+4-8x+4{{y}^{2}})$
Rearranging left hand side and right hand side we get that
$4{{x}^{2}}+4-8x+4{{y}^{2}}-{{x}^{2}}-1-2x-{{y}^{2}}=0$
$\Rightarrow 3{{x}^{2}}+3{{y}^{2}}-10x+3=0$

Dividing 3 on both sides, we get :
${{x}^{2}}+{{y}^{2}}-\dfrac{10}{3}x+1=0$
Compare it with the equation of circle, i.e:
${{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ 2g}x\text{ }+\text{ 2f}y\text{ }+\text{ c }=\text{ }0,$where (-g , -f) is centre of circle
With this we get that $g=-\dfrac{5}{3}$ and $f=0$
Therefore, the centre of the circle is $\left( \dfrac{5}{3},0 \right)$

Hence, the correct option is D.

Note: The general equation of circle is ${{(x-a)}^{2}}+{{(y-b)}^{2}}=r$
Where (a,b) represents the center of the circle, and r is the radius of the circle. If we now shift the origin to the point A, the equation becomes ${{(x)}^{2}}+{{(y)}^{2}}=r$
Complex numbers are numbers that consist of two parts, real numbers and an imaginary number. Complex numbers are the building blocks of more intricate math, such as Algebra.