`zbar(z) + ( 3-i) z+ ( 3+i) bar(z) + 1 =0` represent a circle with

Options

A. Center `( -3,-1)` and radius 3
B. Center `( -3,1)` and radius 3
C. Center `( -3,-1)` and radius 4
D. Center `( -3,1)` and radius 4

Answer

To solve the equation \( \overline{z}(z) + (3-i)z + (3+i)\overline{z} + 1 = 0 \) and determine the center and radius of the circle it represents, we will follow these steps:

### Step 1: Substitute \( z \) and \( \overline{z} \)
Assume \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then, the conjugate \( \overline{z} = x - iy \).

### Step 2: Expand \( \overline{z}(z) \)
Calculate \( \overline{z}(z) \):
\[ \overline{z}(z) = (x - iy)(x + iy) = x^2 + y^2 \]

### Step 3: Expand the equation
Substituting \( z \) and \( \overline{z} \) into the equation:
\[ x^2 + y^2 + (3-i)(x + iy) + (3+i)(x - iy) + 1 = 0 \]

### Step 4: Expand \( (3-i)(z) \) and \( (3+i)(\overline{z}) \)
Calculating \( (3-i)(z) \):
\[ (3-i)(x + iy) = 3x + 3iy - ix - y = (3x - y) + i(3y - x) \]
Calculating \( (3+i)(\overline{z}) \):
\[ (3+i)(x - iy) = 3x - 3iy + ix - y = (3x - y) + i(-3y + x) \]

### Step 5: Combine real and imaginary parts
Now combine all parts:
\[ x^2 + y^2 + (3x - y) + (3x - y) + 1 = 0 \]
The imaginary parts cancel out, so we only consider the real part:
\[ x^2 + y^2 + 6x + 1 = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 + 6x + 1 = 0 \]
To express it in standard form, we complete the square for \( x \):
\[ (x^2 + 6x + 9) + y^2 - 9 + 1 = 0 \]
This simplifies to:
\[ (x + 3)^2 + y^2 - 8 = 0 \]
or:
\[ (x + 3)^2 + y^2 = 8 \]

### Step 7: Identify center and radius
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \):
- The center \( (h, k) = (-3, 0) \)
- The radius \( r = \sqrt{8} = 2\sqrt{2} \)

### Final Answer
The equation represents a circle with:
- Center: \( (-3, 0) \)
- Radius: \( 2\sqrt{2} \)