If the equation `kx^(2) + (2k - 3) y^(2) - 6x + 4y + 3 = 0 ` represents a circle, then its centre is

Options

A. `( - 1, (2)/(3))`
B. ` (2,-(4)/(3))`
C. ` (1,-(2)/(3))`
D. ` (1, (2)/(3))`

Answer

To determine the center of the circle represented by the equation \( kx^2 + (2k - 3)y^2 - 6x + 4y + 3 = 0 \), we need to follow these steps:

### Step 1: Identify the conditions for the equation to represent a circle
For the given equation to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal, and the coefficient of \( xy \) must be zero.

### Step 2: Set the coefficients equal
The coefficient of \( x^2 \) is \( k \) and the coefficient of \( y^2 \) is \( 2k - 3 \). Therefore, we set:
\[ k = 2k - 3 \]

### Step 3: Solve for \( k \)
Rearranging the equation:
\[ k - 2k = -3 \implies -k = -3 \implies k = 3 \]

### Step 4: Substitute \( k \) back into the equation
Now, substitute \( k = 3 \) into the original equation:
\[ 3x^2 + (2(3) - 3)y^2 - 6x + 4y + 3 = 0 \]
This simplifies to:
\[ 3x^2 + 3y^2 - 6x + 4y + 3 = 0 \]

### Step 5: Divide the entire equation by 3
To simplify the equation, divide everything by 3:
\[ x^2 + y^2 - 2x + \frac{4}{3}y + 1 = 0 \]

### Step 6: Rearrange into standard form
Rearranging gives:
\[ x^2 - 2x + y^2 + \frac{4}{3}y + 1 = 0 \]

### Step 7: Complete the square for \( x \) and \( y \)
For \( x \):
\[ x^2 - 2x \rightarrow (x - 1)^2 - 1 \]
For \( y \):
\[ y^2 + \frac{4}{3}y \rightarrow \left(y + \frac{2}{3}\right)^2 - \frac{4}{9} \]

### Step 8: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ (x - 1)^2 - 1 + \left(y + \frac{2}{3}\right)^2 - \frac{4}{9} + 1 = 0 \]
Simplifying this results in:
\[ (x - 1)^2 + \left(y + \frac{2}{3}\right)^2 = \frac{4}{9} \]

### Step 9: Identify the center
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify the center \((h, k)\):
\[ \text{Center} = (1, -\frac{2}{3}) \]

### Final Answer
The center of the circle is:
\[ \boxed{(1, -\frac{2}{3})} \]