Find the coordinates of the centres and the radii of the circles whose equations are
`3x ^(2) + 3y ^(2) - 5x - 6y + 4 = 0`

Answer

To find the coordinates of the center and the radius of the circle given by the equation \( 3x^2 + 3y^2 - 5x - 6y + 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 3x^2 + 3y^2 - 5x - 6y + 4 = 0 \]
To simplify, we can divide the entire equation by 3:
\[ x^2 + y^2 - \frac{5}{3}x - 2y + \frac{4}{3} = 0 \]

### Step 2: Rearrange the equation
We rearrange the equation to group the \(x\) and \(y\) terms:
\[ x^2 - \frac{5}{3}x + y^2 - 2y = -\frac{4}{3} \]

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

**For \(x\):**
1. Take the coefficient of \(x\), which is \(-\frac{5}{3}\), halve it to get \(-\frac{5}{6}\), and square it to get \(\left(-\frac{5}{6}\right)^2 = \frac{25}{36}\).
2. Add and subtract \(\frac{25}{36}\) inside the equation.

**For \(y\):**
1. Take the coefficient of \(y\), which is \(-2\), halve it to get \(-1\), and square it to get \((-1)^2 = 1\).
2. Add and subtract \(1\) inside the equation.

So we rewrite the equation as:
\[ \left(x - \frac{5}{6}\right)^2 - \frac{25}{36} + \left(y - 1\right)^2 - 1 = -\frac{4}{3} \]

### Step 4: Simplify the equation
Now, we simplify the equation:
\[ \left(x - \frac{5}{6}\right)^2 + \left(y - 1\right)^2 = -\frac{4}{3} + \frac{25}{36} + 1 \]
We need a common denominator to combine the right side:
\[ -\frac{4}{3} = -\frac{48}{36}, \quad 1 = \frac{36}{36} \]
Thus,
\[ -\frac{48}{36} + \frac{25}{36} + \frac{36}{36} = \frac{13}{36} \]

### Step 5: Write the standard form of the circle's equation
Now we have:
\[ \left(x - \frac{5}{6}\right)^2 + \left(y - 1\right)^2 = \frac{13}{36} \]
This is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

### Step 6: Identify the center and radius
From the equation, we can identify:
- The center \((h, k) = \left(\frac{5}{6}, 1\right)\)
- The radius \(r = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6}\)

### Final Answer
- **Center:** \(\left(\frac{5}{6}, 1\right)\)
- **Radius:** \(\frac{\sqrt{13}}{6}\)