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

Options

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

Explanation

To solve the equation \( \overline{z} z + (3 - i)z + (3 + i)\overline{z} + 1 = 0 \) and show that it represents a circle, we can follow these steps:

### Step 1: Substitute \( z \) with \( x + iy \)
Let \( z = x + iy \) where \( x \) and \( y \) are real numbers. The conjugate \( \overline{z} \) will be \( x - iy \).

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

### Step 3: Expand the terms
First, expand \( (x - iy)(x + iy) \):
\[ (x - iy)(x + iy) = x^2 + y^2 \]
Next, expand \( (3 - i)(x + iy) \):
\[ (3 - i)(x + iy) = 3x + 3iy - ix - y = (3x - y) + i(3y - x) \]
Now expand \( (3 + i)(x - iy) \):
\[ (3 + i)(x - iy) = 3x - 3iy + ix - y = (3x - y) + i(-3y + x) \]

### Step 4: Combine all parts
Combining all the parts, we have:
\[ x^2 + y^2 + (3x - y) + i(3y - x) + (3x - y) + i(-3y + x) + 1 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 6x + 1 + 0 = 0 \]
The imaginary parts cancel out.

### Step 5: Rearrange the equation
Rearranging gives:
\[ x^2 + y^2 + 6x + 1 = 0 \]

### Step 6: Complete the square
To complete the square for \( x \):
\[ x^2 + 6x = (x + 3)^2 - 9 \]
Substituting this back into the equation:
\[ (x + 3)^2 - 9 + y^2 + 1 = 0 \]
This simplifies to:
\[ (x + 3)^2 + y^2 - 8 = 0 \]
or
\[ (x + 3)^2 + y^2 = 8 \]

### Step 7: Identify the center and radius
This is the equation of a circle in standard form:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( h = -3 \), \( k = 0 \), and \( r^2 = 8 \). Thus, the center is \( (-3, 0) \) and the radius is \( \sqrt{8} = 2\sqrt{2} \).

### Final Answer
The equation represents a circle with center \( (-3, 0) \) and radius \( 2\sqrt{2} \).

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