What is the center and radius of a circle whose equation is `3x^(2)+3y^(2)-12x+18y=69`?

Answer

To find the center and radius of the circle given by the equation \(3x^2 + 3y^2 - 12x + 18y = 69\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 3x^2 + 3y^2 - 12x + 18y = 69 \]
To simplify, we can divide the entire equation by 3:
\[ x^2 + y^2 - 4x + 6y = 23 \]

### Step 2: Rearrange the equation
Now, we rearrange the equation to group the \(x\) and \(y\) terms:
\[ x^2 - 4x + y^2 + 6y = 23 \]

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

For \(x^2 - 4x\):
- Take half of \(-4\), which is \(-2\), and square it: \((-2)^2 = 4\).
- Add and subtract 4:
\[ x^2 - 4x + 4 - 4 \]

For \(y^2 + 6y\):
- Take half of \(6\), which is \(3\), and square it: \(3^2 = 9\).
- Add and subtract 9:
\[ y^2 + 6y + 9 - 9 \]

Now, substituting these into the equation:
\[ (x^2 - 4x + 4) + (y^2 + 6y + 9) - 4 - 9 = 23 \]
This simplifies to:
\[ (x - 2)^2 + (y + 3)^2 - 13 = 23 \]

### Step 4: Simplify the equation
Now, we move \(-13\) to the right side:
\[ (x - 2)^2 + (y + 3)^2 = 36 \]

### Step 5: Identify the center and radius
Now, we can identify the center and radius from the standard form of the circle equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

From our equation:
- The center \((h, k)\) is \((2, -3)\).
- The radius \(r\) is \(\sqrt{36} = 6\).

### Final Answer
The center of the circle is \((2, -3)\) and the radius is \(6\).

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