Question

The equation $z\bar z + a\bar z + \bar az + b = 0,b \in R$ represents a circle (not point circle) if
A) ${\left| a \right|^2} > b$
B) ${\left| a \right|^2} < b$
C) $\left| a \right| > b$
D) $\left| a \right| < b$

Answer 1

Here we have to solve the equation to get the required condition. We will first simplify the equation by using the properties of the quadratic equation. Then we will apply the condition of the circle not being a point circle to get the required condition.

Complete step by step solution:
We will simplify the given equation. So we will add $a\bar a$ on the both side of the equation.
Therefore,
$z\bar z + a\bar z + \bar az + b + a\bar a = 0 + a\bar a$
Now taking the common terms, we get
$\Rightarrow z\left( {\bar z + \bar a} \right) + a\left( {\bar z + \bar a} \right) + b = a\bar a$
We know from the properties of the quadratic equations, $a\left( {x + y} \right) + b\left( {x + y} \right) = \left( {a + b} \right)\left( {x + y} \right)$. Therefore, we get
$\Rightarrow \left( {z + a} \right)\left( {\bar z + \bar a} \right) = a\bar a - b$
Now we know that $x.\bar x = {\left| x \right|^2}$, then by using this property ,we get
$\Rightarrow {\left| {z + a} \right|^2} = {\left| a \right|^2} - b$
It is given in the question that the circle is not the point circle. Therefore the value of ${\left| {z + a} \right|^2}$ must not be equal to zero. Therefore, we get
${\left| {z + a} \right|^2} > 0$
Substituting ${\left| {z + a} \right|^2} = {\left| a \right|^2} - b$ in the above inequation, we get
$\Rightarrow {\left| a \right|^2} - b > 0$
$\Rightarrow {\left| a \right|^2} > b$

So, option A is the correct option.

Note:
Here it is given that the circle is not a point circle. Point circle means the radius is equal to 0. So, if the circle is not a point circle, then the radius must be greater than zero. If we substitute radius equals to 0 then we will get ${\left| a \right|^2} = b$, which will be wrong. In addition to this, $\bar x$ means that it is the complementary function of $x$. We need to keep in mind $x.\bar x = {\left| x \right|^2}$ means multiplication of the function to its complementary function is equal to its magnitude square.