Length of a tangent drawn from the origin to the circle `x ^(2) + y ^(2) - 6x + 4y + 8=0 ` in units is

Options

A. 4
B. 6
C. 8
D. `2sqrt2`

Answer

To find the length of the tangent drawn from the origin to the given circle, we can follow these steps:

### Step 1: Write the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 4y + 8 = 0 \]

We can rearrange this equation to identify the center and radius of the circle. We will complete the square for both \(x\) and \(y\).

### Step 2: Complete the square for \(x\) and \(y\)
1. For \(x\):
\[ x^2 - 6x \quad \text{can be written as} \quad (x - 3)^2 - 9 \]

2. For \(y\):
\[ y^2 + 4y \quad \text{can be written as} \quad (y + 2)^2 - 4 \]

Now substituting these back into the circle equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 + 8 = 0 \]
Simplifying this gives:
\[ (x - 3)^2 + (y + 2)^2 - 5 = 0 \]
Thus, we have:
\[ (x - 3)^2 + (y + 2)^2 = 5 \]

### Step 3: Identify the center and radius of the circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (3, -2)\)
- Radius \(r = \sqrt{5}\)

### Step 4: Use the formula for the length of the tangent
The length of the tangent \(L\) from a point \((x_1, y_1)\) to a circle \((x - h)^2 + (y - k)^2 = r^2\) is given by:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

### Step 5: Substitute the values
Here, the point is the origin \((0, 0)\), so:
- \(x_1 = 0\), \(y_1 = 0\)
- \(h = 3\), \(k = -2\), \(r = \sqrt{5}\)

Now substituting these values into the formula:
\[ L = \sqrt{(0 - 3)^2 + (0 + 2)^2 - (\sqrt{5})^2} \]
Calculating each term:
\[ L = \sqrt{(-3)^2 + (2)^2 - 5} \]
\[ L = \sqrt{9 + 4 - 5} \]
\[ L = \sqrt{8} \]
\[ L = 2\sqrt{2} \]

### Final Answer
The length of the tangent drawn from the origin to the circle is:
\[ \boxed{2\sqrt{2}} \]