Find the radius and centre of the circle
`z bar(z)- (2 + 3i) z-(2-3i) bar(z)+ 9=0` where z is a complex variable

Answer

To find the radius and center of the circle given by the equation:

\[ \overline{z} z - (2 + 3i) z - (2 - 3i) \overline{z} + 9 = 0 \]

where \( z \) is a complex variable, we will follow these steps:

### Step 1: Substitute \( z \) and \( \overline{z} \)

Let \( z = x + yi \) and \( \overline{z} = x - yi \). Substitute these into the equation:

\[ (x - yi)(x + yi) - (2 + 3i)(x + yi) - (2 - 3i)(x - yi) + 9 = 0 \]

### Step 2: Simplify \( z \overline{z} \)

Using the identity \( z \overline{z} = |z|^2 = x^2 + y^2 \):

\[ x^2 + y^2 - (2 + 3i)(x + yi) - (2 - 3i)(x - yi) + 9 = 0 \]

### Step 3: Expand the terms

Now, expand the terms involving \( z \) and \( \overline{z} \):

1. \( (2 + 3i)(x + yi) = 2x + 2yi + 3xi - 3y = (2x - 3y) + (2y + 3x)i \)
2. \( (2 - 3i)(x - yi) = 2x - 2yi - 3xi - 3y = (2x + 3y) + (-2y + 3x)i \)

Now substitute these back into the equation:

\[ x^2 + y^2 - [(2x - 3y) + (2y + 3x)i] - [(2x + 3y) + (-2y + 3x)i] + 9 = 0 \]

### Step 4: Combine like terms

Combine the real and imaginary parts:

Real part:
\[ x^2 + y^2 - (2x - 3y) - (2x + 3y) + 9 = 0 \]
\[ x^2 + y^2 - 4x + 6y + 9 = 0 \]

Imaginary part:
\[ -(2y + 3x) - (-2y + 3x) = 0 \]

### Step 5: Rearrange the equation

Now, rearranging the real part gives us:

\[ x^2 + y^2 - 4x + 6y + 9 = 0 \]

### Step 6: Complete the square

To express this in the standard form of a circle, we complete the square for \( x \) and \( y \):

1. For \( x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

2. For \( y \):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting these into the equation gives:

\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 + 9 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 4 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 4 \]

### Step 7: Identify the center and radius

From the equation \( (x - 2)^2 + (y + 3)^2 = 4 \), we can identify:

- The center of the circle is \( (2, -3) \).
- The radius of the circle is \( \sqrt{4} = 2 \).

### Final Answer

- **Center**: \( (2, -3) \)
- **Radius**: \( 2 \)