Find the area of the equilateral triangle that can be inscribed in the circle :
`x^(2) + y^(2) -4x + 6y - 3 = 0 `

Answer

To find the area of the equilateral triangle that can be inscribed in the given circle, we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 6y - 3 = 0 \]

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

1. For \(x\):
\[ x^2 - 4x \quad \text{can be rewritten as} \quad (x - 2)^2 - 4 \]

2. For \(y\):
\[ y^2 + 6y \quad \text{can be rewritten as} \quad (y + 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \]
Simplifying this:
\[ (x - 2)^2 + (y + 3)^2 - 16 = 0 \]
Thus, we have:
\[ (x - 2)^2 + (y + 3)^2 = 16 \]

### Step 2: Identify the center and radius of the circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can see that:
- The center of the circle \((h, k)\) is \((2, -3)\).
- The radius \(r\) is \(\sqrt{16} = 4\).

### Step 3: Find the side length of the inscribed equilateral triangle
The side length \(a\) of an equilateral triangle inscribed in a circle of radius \(r\) can be calculated using the formula:
\[ a = r \cdot \sqrt{3} \]
Substituting \(r = 4\):
\[ a = 4 \cdot \sqrt{3} = 4\sqrt{3} \]

### Step 4: Calculate the area of the equilateral triangle
The area \(A\) of an equilateral triangle with side length \(a\) is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
Substituting \(a = 4\sqrt{3}\):
\[ A = \frac{\sqrt{3}}{4} (4\sqrt{3})^2 \]
Calculating \( (4\sqrt{3})^2 \):
\[ (4\sqrt{3})^2 = 16 \cdot 3 = 48 \]
Thus:
\[ A = \frac{\sqrt{3}}{4} \cdot 48 = 12\sqrt{3} \]

### Final Answer
The area of the equilateral triangle that can be inscribed in the circle is:
\[ \boxed{12\sqrt{3}} \]