Question

If $\dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2$ represents a circle, then its radius is equal to
A. $1$
B. $\dfrac{1}{3}$
C. $\dfrac{3}{4}$
D. $\dfrac{2}{3}$

Answer 1

Hint: In this question, we have to find the radius of the circle that is obtained from the given equation. To do this, the standard form of a complex number $z=x+iy$ is substituted in the given equation, and on simplifying, we get the equation of a circle. From this, we can able to extract the radius using the appropriate formula.

Formula Used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$
The standard form of a circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ whose centre is at $(-g,-f)$ and its radius is $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.

Complete step by step solution: Given equation is
$\dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2$
Consider the complex number $z=x+iy$
Substituting in the given equation, we get
$\begin{align}
  & \dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2 \\
 & \Rightarrow \left| z-2 \right|=2\left| z-3 \right| \\
 & \Rightarrow \left| (x+iy)-2 \right|=2\left| (x+iy)-3 \right| \\
 & \Rightarrow \left| (x-2)+iy \right|=2\left| (x-3)+iy \right| \\
\end{align}$
$\Rightarrow \sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}$
Squaring on both sides, we get
$\begin{align}
  & \Rightarrow {{(x-2)}^{2}}+{{y}^{2}}=4\left( {{(x-3)}^{2}}+{{y}^{2}} \right) \\
 & \Rightarrow {{x}^{2}}-2x+4+{{y}^{2}}=4({{x}^{2}}-6x+9+{{y}^{2}}) \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2x+4=4{{x}^{2}}+4{{y}^{2}}-24x+36 \\
 & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}-20x+32=0 \\
\end{align}$
Since the standard form of the circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, we can write the obtained equation as
${{x}^{2}}+{{y}^{2}}-\dfrac{20}{3}x+\dfrac{32}{3}=0$
So, on comparing, we get
$\begin{align}
  & 2g=\dfrac{-20}{3} \\
 & \Rightarrow g=\dfrac{-10}{3} \\
 & f=0;c=\dfrac{32}{3} \\
\end{align}$
So, the radius of the circle is
\[\begin{align}
  & r=\sqrt{{{g}^{2}}+{{f}^{2}}-c} \\
 & \text{ }=\sqrt{{{\left( \dfrac{-10}{3} \right)}^{2}}+0-\left( \dfrac{32}{3} \right)} \\
 & \text{ }=\sqrt{\dfrac{100}{9}-\dfrac{32}{3}} \\
 & \text{ }=\sqrt{\dfrac{100-96}{9}} \\
 & \text{ }=\sqrt{\dfrac{4}{9}} \\
 & \text{ }=\dfrac{2}{3} \\
\end{align}\]

Option ‘D’ is correct

Note: Here we need to substitue the complex number in place of $z$ in the given equation. So, that we can able to find the magnitude of the complex numbers and extract the equation of the circle. Therefore, we can calculate the radius by comparing the obtained equation with the standard form of the circle.