Question

A circle whose radius is $r$ and centre ${{z}_{0}}$, then the equation of the circle is
A. $z\overline{z}-z\overline{{{z}_{0}}}-\overline{z}{{z}_{0}}+{{z}_{0}}\overline{{{z}_{0}}}={{r}^{2}}$
B. $z\overline{z}+z\overline{{{z}_{0}}}-\overline{z}{{z}_{0}}+{{z}_{0}}\overline{{{z}_{0}}}={{r}^{2}}$
C. $z\overline{z}-z\overline{{{z}_{0}}}+\overline{z}{{z}_{0}}-{{z}_{0}}\overline{{{z}_{0}}}={{r}^{2}}$
D. None of these

Answer 1

Hint: In this question, we have to find the equation of the circle whose centre and radius are ${{z}_{0}}$ and $r$ respectively. To find this, the standard form of a circle is used.

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 $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: Given that,
A circle has centre at ${{z}_{0}}({{x}_{0}},{{y}_{0}})$ i.e., ${{z}_{0}}={{x}_{0}}+i{{y}_{0}}$ and radius $r$.
The equation of a circle is
${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$ where the centre is $(h,k)$ and radius is $r$.
So, the given circle’s equation is
${{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}={{r}^{2}}\text{ }...(1)$
But we have
$\begin{align}
  & z=x+iy=(x,y) \\
 & {{z}_{0}}={{x}_{0}}+i{{y}_{0}}=({{x}_{0}},{{y}_{0}}) \\
\end{align}$
So, we can write
$\begin{align}
  & \left| z-{{z}_{0}} \right|=\left| (x+iy)-({{x}_{0}}+i{{y}_{0}}) \right| \\
 & \text{ }=\left| (x-{{x}_{0}})+i(y-{{y}_{0}}) \right| \\
\end{align}$
On squaring, we get
$\begin{align}
  & {{\left| z-{{z}_{0}} \right|}^{2}}={{\left| (x-{{x}_{0}})+i(y-{{y}_{0}}) \right|}^{2}} \\
 & \text{ }={{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}\text{ }...(2) \\
\end{align}$
Then, from (1) and (2),
\[\begin{align}
  & {{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}={{r}^{2}} \\
 & \Rightarrow {{\left| z-{{z}_{0}} \right|}^{2}}={{r}^{2}} \\
 & \Rightarrow (z-{{z}_{0}})(\overline{z-{{z}_{0}}})={{r}^{2}} \\
 & \Rightarrow (z-{{z}_{0}})(\overline{z}-\overline{{{z}_{0}}})={{r}^{2}} \\
 & \Rightarrow z\overline{z}-z\overline{{{z}_{0}}}-{{z}_{0}}\overline{z}+{{z}_{0}}\overline{{{z}_{0}}}={{r}^{2}} \\
\end{align}\]

Option ‘A’ is correct

Note: Here we need to use the general formula for the equation of a circle with the centre and radius. Then, we can able to find the required equation.