Question

The equation \[z\overline{z}+(2-3i)z+(2+3i)\overline{z}+4=0\] represents a circle of radius
A. $2$
B. $3$
C. $4$
D. $6$

Answer 1

Hint: In this question, we have to find the radius of the circle formed by the given equation. Here the general form of a complex number and its conjugate are substituted in the given equation, to find the required radius.

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, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
The complex conjugate is represented by $\overline{z}=x-iy$.
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: Consider a complex number represented by $z=x+iy$ and its conjugate is represented by $\overline{z}=x-iy$.
The given equation is
\[z\overline{z}+(2-3i)z+(2+3i)\overline{z}+4=0\]
On substituting the complex number and its conjugate in the given equation, we get
\[\begin{align}
  & z\overline{z}+(2-3i)z+(2+3i)\overline{z}+4=0 \\
 & \Rightarrow (x+iy)(x-iy)+(2-3i)(x+iy)+(2+3i)(x-iy)+4=0 \\
 & \Rightarrow {{x}^{2}}-{{i}^{2}}{{y}^{2}}+2x+i2y-i3x-{{i}^{2}}3y+2x-i2y+i3x-{{i}^{2}}3y+4=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+4x+6y+4=0 \\
\end{align}\]
Where ${{i}^{2}}=-1$;
Thus, the obtained equation is ${{x}^{2}}+{{y}^{2}}+4x+6y+4=0$ is familiar with the equation of a circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. So, the radius of the circle is calculated by $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
Here $g=2;f=3;c=4$
On substituting for the radius, we get
$\begin{align}
  & r=\sqrt{{{g}^{2}}+{{f}^{2}}-c} \\
 & \text{ }=\sqrt{{{2}^{2}}+{{3}^{2}}-4} \\
 & \text{ }=\sqrt{4+9-4} \\
 & \text{ }=3 \\
\end{align}$

Option ‘B’ is correct

Note: Here we need to compare the obtained equation by the substitution of the complex number and its conjugate with the general equation of a circle. So, we can able to calculate the radius.