Find the centre and radius of the circle `x^(2) + y^(2) - 6x + 4y - 12 =0`.

Explanation

To find the center and radius of the circle given by the equation \(x^2 + y^2 - 6x + 4y - 12 = 0\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of a circle's equation is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. We start by rearranging the given equation:
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