The equation `zbarz+abarz+baraz+b=0, b in R` represents circle, if

Options

A. `|a|^(2) =b`
B. `|a|^(2) gt b`
C. `|a|^(2) lt b`
D. none of these

Explanation

To determine the conditions under which the equation \( z \bar{z} + \alpha \bar{z} + \bar{\alpha} z + b = 0 \) represents a circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Equation**:
We start with the equation:
\[ z \bar{z} + \alpha \bar{z} + \bar{\alpha} z + b = 0 \]
Here, \( z \) is a complex number, and \( \alpha \) is also a complex number.

2. **Rewrite the Equation**:
We can rewrite the equation in a more standard form. Recall that \( z \bar{z} = |z|^2 \). Thus, the equation can be expressed as:
\[ |z|^2 + \alpha \bar{z} + \bar{\alpha} z + b = 0 \]

3. **Standard Form of Circle**:
The standard form of a circle in the complex plane is:
\[ |z - z_0|^2 = r^2 \]
where \( z_0 \) is the center and \( r \) is the radius.

4. **Identify the Center and Radius**:
From the equation, we can identify the center and radius. The term \( \alpha \bar{z} + \bar{\alpha} z \) can be rearranged to express the center. The center \( z_0 \) can be found as:
\[ z_0 = -\frac{\alpha}{2} \]
The radius \( r \) can be derived from the constant term \( b \):
\[ r^2 = |\alpha|^2 - b \]

5. **Condition for Circle**:
For the equation to represent a circle, the radius must be a positive real number:
\[ r^2 > 0 \implies |\alpha|^2 - b > 0 \implies |\alpha|^2 > b \]

### Conclusion:
Thus, the equation \( z \bar{z} + \alpha \bar{z} + \bar{\alpha} z + b = 0 \) represents a circle if:
\[ |\alpha|^2 > b \]