Fill up in each of the following
(a) the centre of the circle `3x^2+3y^2+6x-12y-6=0` is ______

Answer

To find the center of the circle given by the equation \(3x^2 + 3y^2 + 6x - 12y - 6 = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of a circle's equation is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
In our case, we have coefficients in front of \(x^2\) and \(y^2\). To simplify, we will divide the entire equation by 3.

**Calculation:**
\[ \frac{3x^2 + 3y^2 + 6x - 12y - 6}{3} = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x - 4y - 2 = 0 \]

### Step 2: Identify the coefficients
Now, we can compare our simplified equation with the standard form. From the equation \(x^2 + y^2 + 2x - 4y - 2 = 0\), we can identify:
- \(2g = 2\) which gives \(g = 1\)
- \(2f = -4\) which gives \(f = -2\)
- \(c = -2\)

### Step 3: Find the center of the circle
The center of the circle \((h, k)\) can be found using the values of \(g\) and \(f\):
\[ h = -g = -1 \]
\[ k = -f = 2 \]

Thus, the center of the circle is:
\[ (-1, 2) \]

### Final Answer:
The center of the circle is \((-1, 2)\).

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