Fill up in each of the following
(b) the radius of the circle is `3x^2+3y^2+6x-12y-15=0`

Answer

To find the radius of the circle given by the equation \(3x^2 + 3y^2 + 6x - 12y - 15 = 0\), 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 the given equation, the coefficients of \(x^2\) and \(y^2\) are both 3. To simplify, we can divide the entire equation by 3.

\[ \frac{3x^2}{3} + \frac{3y^2}{3} + \frac{6x}{3} - \frac{12y}{3} - \frac{15}{3} = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x - 4y - 5 = 0 \]

### Step 2: Identify coefficients
Now, we can identify the coefficients \(g\), \(f\), and \(c\) from the standard form:
- \(g = 1\) (coefficient of \(x\))
- \(f = -2\) (coefficient of \(y\))
- \(c = -5\)

### Step 3: Use the radius formula
The radius \(r\) of the circle can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values we found:
\[ r = \sqrt{1^2 + (-2)^2 - (-5)} \]

### Step 4: Calculate the radius
Now, calculate the values:
- \(g^2 = 1^2 = 1\)
- \(f^2 = (-2)^2 = 4\)
- \(-c = -(-5) = 5\)

Putting it all together:
\[ r = \sqrt{1 + 4 + 5} = \sqrt{10} \]

### Final Answer
Thus, the radius of the circle is:
\[ \sqrt{10} \text{ units} \]