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 can follow these steps:

### Step 1: Rewrite the Circle Equation
We start by rewriting the circle equation in standard form. The given equation is:

\[ x^2 + y^2 + 4x + 6y + 12 = 0 \]

We complete the square for both \(x\) and \(y\).

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

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

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 is:

\[ (x + 2)^2 + (y + 3)^2 = 1 \]

### Step 2: Identify the Center and Radius
From the standard form of the circle, we can identify:

- Center \(C\) is \((-2, -3)\)
- Radius \(r = 1\)

### Step 3: Calculate the Distance from the Origin to the Center
Next, we calculate the distance \(OC\) from the origin \(O(0, 0)\) to the center \(C(-2, -3)\):

\[ OC = \sqrt{(-2 - 0)^2 + (-3 - 0)^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 4: Use the Circumradius Formula
The circumradius \(R\) of triangle \(OAB\) can be calculated using the formula:

\[ R = \frac{OC}{2} \]

Substituting the value of \(OC\):

\[ R = \frac{\sqrt{13}}{2} \]

### Conclusion
Thus, the circumradius of triangle \(OAB\) is:

\[ \boxed{\frac{\sqrt{13}}{2}} \]