The radius of the circle `x^(2) + y^(2) + 4x - 6y + 12 = 0` is

Options

A. 5
B. `sqrt(13)`
C. 1
D. `sqrt(17)`

Answer

To find the radius of the circle given by the equation \(x^{2} + y^{2} + 4x - 6y + 12 = 0\), 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 of the circle and \(r\) is the radius.

### Step 2: Rearrange the given equation
Start by rearranging the given equation:
\[ x^{2} + y^{2} + 4x - 6y + 12 = 0 \]
This can be rearranged to:
\[ x^{2} + 4x + y^{2} - 6y + 12 = 0 \]
Now, isolate the constant term:
\[ x^{2} + 4x + y^{2} - 6y = -12 \]

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

**For \(x\):**
\[ x^{2} + 4x \rightarrow (x + 2)^{2} - 4 \]

**For \(y\):**
\[ y^{2} - 6y \rightarrow (y - 3)^{2} - 9 \]

### Step 4: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ ((x + 2)^{2} - 4) + ((y - 3)^{2} - 9) = -12 \]
Simplifying this, we have:
\[ (x + 2)^{2} + (y - 3)^{2} - 13 = -12 \]
\[ (x + 2)^{2} + (y - 3)^{2} = 1 \]

### Step 5: Identify the radius
From the standard form \((x - h)^{2} + (y - k)^{2} = r^{2}\), we see that:
\[ r^{2} = 1 \implies r = \sqrt{1} = 1 \]

### Conclusion
Thus, the radius of the circle is:
\[ \boxed{1} \]