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

Options

A. `|a|^2 = |b|^2`
B. `|a|^2 ge b`
C. `(3 sqrt3)/(4)`
D. None of these

Answer

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

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ z \overline{z} + \alpha \overline{z} + \overline{\alpha} z + b = 0 \]
This can be rearranged to:
\[ |z|^2 + \alpha \overline{z} + \overline{\alpha} z + b = 0 \]

### Step 2: Identify the standard form of a 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. Expanding this gives:
\[ |z|^2 - 2 \text{Re}(z_0 \overline{z}) + |z_0|^2 = r^2 \]
This can be rearranged to:
\[ |z|^2 + b = -\alpha \overline{z} - \overline{\alpha} z \]

### Step 3: Compare coefficients
From the equation \( |z|^2 + \alpha \overline{z} + \overline{\alpha} z + b = 0 \), we can see that:
- The term \( \alpha \overline{z} + \overline{\alpha} z \) corresponds to \( -2 \text{Re}(z_0 \overline{z}) \).
- Thus, we can identify \( z_0 = -\alpha \).

### Step 4: Determine the radius
The radius \( r \) can be found from:
\[ r^2 = |\alpha|^2 - b \]
For the equation to represent a circle, the radius must be non-negative:
\[ |\alpha|^2 - b \geq 0 \]

### Step 5: Rearranging the inequality
This leads to the condition:
\[ |\alpha|^2 \geq b \]

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