Two tangents OA and OB are drawn to the circle `x^(2)+y^(2)+4x+6y+12=0` from origin O. The circumradius of `DeltaOAB` is :

Options

A. `(1)/(2)`
B. 1
C. 2
D. `(sqrt(13))/(2)`

Explanation

To find the circumradius of triangle OAB formed by the tangents OA and OB drawn from the origin O to the circle given by the equation \(x^2 + y^2 + 4x + 6y + 12 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 4x + 6y + 12 = 0 \]
We can rearrange this equation to identify the center and radius of the circle. To do this, we complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 + 4x = (x + 2)^2 - 4 \]
2. For \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 + 12 = 0 \]
Simplifying this, we have:
\[ (x + 2)^2 + (y + 3)^2 - 1 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x + 2)^2 + (y + 3)^2 = 1 \]
From this, we can see that the center of the circle is \((-2, -3)\) and the radius is \(1\).

### Step 2: Find the distance from the origin to the center of the circle
The distance \(OC\) from the origin \(O(0, 0)\) to the center \(C(-2, -3)\) can be calculated using the distance formula:
\[ OC = \sqrt{(-2 - 0)^2 + (-3 - 0)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 3: Calculate the circumradius of triangle OAB
The circumradius \(R\) of triangle OAB, where OA and OB are tangents to the circle, can be calculated using the formula:
\[ R = \frac{OC}{2} \]
Substituting the value of \(OC\):
\[ R = \frac{\sqrt{13}}{2} \]

### Final Answer
The circumradius of triangle OAB is:
\[ \frac{\sqrt{13}}{2} \]