If an equilateral triangle is inscribed in the circle `x^(2)+y^(2)-6x-4y+5=0` then its side is

Options

A. `sqrt(6)`
B. `2sqrt(6)`
C. `3sqrt(6)`
D. `4sqrt(6)`

Answer

To find the side length of an equilateral triangle inscribed in the circle given by the equation \(x^2 + y^2 - 6x - 4y + 5 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the given equation of the circle in standard form. The standard form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Starting with the equation:
\[ x^2 + y^2 - 6x - 4y + 5 = 0 \]
we can rearrange it as:
\[ x^2 - 6x + y^2 - 4y = -5 \]

### Step 2: Complete the Square
Now, we complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 - 4y \rightarrow (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 = -5 \]
Simplifying this, we have:
\[ (x - 3)^2 + (y - 2)^2 - 13 = -5 \]
\[ (x - 3)^2 + (y - 2)^2 = 8 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - 3)^2 + (y - 2)^2 = 8\), we can identify:
- Center \((h, k) = (3, 2)\)
- Radius \(r = \sqrt{8} = 2\sqrt{2}\)

### Step 4: Use the Properties of the Equilateral Triangle
For an equilateral triangle inscribed in a circle, the relationship between the side length \(s\) of the triangle and the radius \(r\) of the circle is given by:
\[ s = r \cdot \sqrt{3} \]

### Step 5: Calculate the Side Length
Substituting the radius we found:
\[ s = 2\sqrt{2} \cdot \sqrt{3} = 2\sqrt{6} \]

### Conclusion
Thus, the side length of the equilateral triangle inscribed in the given circle is:
\[ \boxed{2\sqrt{6}} \]