Find the centre and radius of the circle
`3x ^(2) + (a + 1) y ^(2) + 6x -9y + a + 4 =0`

Answer

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

\[ 3x^2 + (a + 1)y^2 + 6x - 9y + (a + 4) = 0 \]

we will follow these steps:

### Step 1: Identify the coefficients
The general form of a circle's equation is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can rewrite it in a more standard form by factoring out the coefficients of \(x^2\) and \(y^2\).

### Step 2: Make the coefficients of \(x^2\) and \(y^2\) equal to 1
To do this, we divide the entire equation by 3 (the coefficient of \(x^2\)):

\[ x^2 + \frac{(a + 1)}{3}y^2 + 2x - 3y + \frac{(a + 4)}{3} = 0 \]

### Step 3: Equate coefficients
For the equation to represent a circle, the coefficients of \(x^2\) and \(y^2\) must be equal. Therefore:

\[ \frac{(a + 1)}{3} = 1 \]

Multiplying both sides by 3 gives:

\[ a + 1 = 3 \implies a = 2 \]

### Step 4: Substitute \(a\) back into the equation
Now, substituting \(a = 2\) back into the equation:

\[ 3x^2 + (2 + 1)y^2 + 6x - 9y + (2 + 4) = 0 \]

This simplifies to:

\[ 3x^2 + 3y^2 + 6x - 9y + 6 = 0 \]

Dividing the entire equation by 3:

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

### Step 5: Rearranging the equation
Now, we rearrange the equation to group \(x\) and \(y\):

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

### Step 6: Completing the square
We complete the square for \(x\) and \(y\):

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

For \(y\):
\[ y^2 - 3y = (y - \frac{3}{2})^2 - \frac{9}{4} \]

Substituting these back into the equation gives:

\[ (x + 1)^2 - 1 + (y - \frac{3}{2})^2 - \frac{9}{4} + 2 = 0 \]

### Step 7: Simplifying the equation
Combining the constants:

\[ (x + 1)^2 + (y - \frac{3}{2})^2 - 1 - \frac{9}{4} + 2 = 0 \]

This simplifies to:

\[ (x + 1)^2 + (y - \frac{3}{2})^2 - \frac{5}{4} = 0 \]

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

\[ (x + 1)^2 + (y - \frac{3}{2})^2 = \frac{5}{4} \]

### Step 9: Identify the center and radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we identify:

- Center \((h, k) = (-1, \frac{3}{2})\)
- Radius \(r = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\)

### Final Answer:
- Center: \((-1, \frac{3}{2})\)
- Radius: \(\frac{\sqrt{5}}{2}\)

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