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

Answer

To find the center and radius of the circle given by the equation \(3x^2 + 3y^2 - 6x + 4y - 4 = 0\), we will follow these steps:

### Step 1: Simplify the equation
We start with the given equation:
\[ 3x^2 + 3y^2 - 6x + 4y - 4 = 0 \]
To make the coefficients of \(x^2\) and \(y^2\) equal to 1, we divide the entire equation by 3:
\[ x^2 + y^2 - 2x + \frac{4}{3}y - \frac{4}{3} = 0 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 + y^2 - 2x + \frac{4}{3}y = \frac{4}{3} \]

### Step 3: Complete the square for \(x\) and \(y\)
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 - 2x \quad \text{can be written as} \quad (x - 1)^2 - 1 \]

For \(y\):
\[ y^2 + \frac{4}{3}y \quad \text{can be written as} \quad \left(y + \frac{2}{3}\right)^2 - \frac{4}{9} \]

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

Combining the constants on the left side:
\[ (x - 1)^2 + \left(y + \frac{2}{3}\right)^2 = \frac{4}{3} + 1 + \frac{4}{9} \]
To combine these, we convert \(1\) to a fraction with a denominator of \(9\):
\[ 1 = \frac{9}{9} \]
Thus:
\[ \frac{4}{3} = \frac{12}{9} \]
Adding these gives:
\[ \frac{12}{9} + \frac{9}{9} + \frac{4}{9} = \frac{25}{9} \]

### Step 5: Write the equation in standard form
Now we can write the equation of the circle in standard form:
\[ (x - 1)^2 + \left(y + \frac{2}{3}\right)^2 = \frac{25}{9} \]

### Step 6: Identify the center and radius
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- The center \((h, k)\) is \((1, -\frac{2}{3})\)
- The radius \(r\) is \(\sqrt{\frac{25}{9}} = \frac{5}{3}\)

### Final Answer
Thus, the center of the circle is \((1, -\frac{2}{3})\) and the radius is \(\frac{5}{3}\).

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