The equation ` x^(2) + y^(2) + 4x - 6y= 0` can be written in the form

Answer

To rewrite the equation \( x^2 + y^2 + 4x - 6y = 0 \) in the standard form of a circle, we will complete the square for both the \( x \) and \( y \) terms. Here are the steps:

### Step 1: Rearrange the equation
Start with the original equation:
\[ x^2 + y^2 + 4x - 6y = 0 \]
Rearranging gives:
\[ x^2 + 4x + y^2 - 6y = 0 \]

### Step 2: Complete the square for the \( x \) terms
Take the \( x \) terms \( x^2 + 4x \):
1. Take half of the coefficient of \( x \) (which is 4), square it: \( \left(\frac{4}{2}\right)^2 = 4 \).
2. Add and subtract this square inside the equation:
\[ x^2 + 4x + 4 - 4 \]
This can be rewritten as:
\[ (x + 2)^2 - 4 \]

### Step 3: Complete the square for the \( y \) terms
Now take the \( y \) terms \( y^2 - 6y \):
1. Take half of the coefficient of \( y \) (which is -6), square it: \( \left(\frac{-6}{2}\right)^2 = 9 \).
2. Add and subtract this square inside the equation:
\[ y^2 - 6y + 9 - 9 \]
This can be rewritten as:
\[ (y - 3)^2 - 9 \]

### Step 4: Substitute back into the equation
Now substitute the completed squares back into the equation:
\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 = 0 \]
Combine the constants:
\[ (x + 2)^2 + (y - 3)^2 - 13 = 0 \]
Rearranging gives:
\[ (x + 2)^2 + (y - 3)^2 = 13 \]

### Step 5: Identify the standard form
The equation is now in the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( h = -2 \), \( k = 3 \), and \( r^2 = 13 \). Thus, the center of the circle is \( (-2, 3) \) and the radius is \( \sqrt{13} \).

### Final Answer
The equation \( x^2 + y^2 + 4x - 6y = 0 \) can be written in the form:
\[ (x + 2)^2 + (y - 3)^2 = 13 \]