The centre of the circle `z bar(z) - ( 2+ 3i) z - ( 2-3i) bar(z) + 9=0`

Options

A. `( 2,-3) `
B. `(2,3)`
C. `(-2,-3)`
D. `(-2,3)`

Answer

To find the center of the circle given by the equation \( \overline{z} z - (2 + 3i) z - (2 - 3i) \overline{z} + 9 = 0 \), we will follow these steps:

### Step 1: Substitute \( z \) and \( \overline{z} \)
Let \( z = x + iy \) and \( \overline{z} = x - iy \). Substitute these into the equation:

\[ \overline{z} z = (x - iy)(x + iy) = x^2 + y^2 \]

### Step 2: Expand the equation
Now substitute \( z \) and \( \overline{z} \) into the original equation:

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

### Step 3: Distribute the terms
Distributing the terms gives:

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

This simplifies to:

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

### Step 4: Combine like terms
Combine the real and imaginary parts:

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

### Step 5: Rearrange the equation
Rearranging gives:

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

### Step 6: Complete the square
To find the center, we need to complete the square for both \( x \) and \( y \):

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

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

### Step 7: Substitute back into the equation
Substituting back, we have:

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

This simplifies to:

\[ (x - 2)^2 + (y + 3)^2 - 13 = -9 \]

### Step 8: Final form of the equation
Rearranging gives:

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

### Step 9: Identify the center
From the equation of the circle \( (x - h)^2 + (y - k)^2 = r^2 \), we can identify the center \( (h, k) \):

The center is \( (2, -3) \).

### Final Answer
The center of the circle is \( (2, -3) \).

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