If `x ^(2) + y ^(2) + 6x + 5 =4 (x-y)` then x-y is

Options

A. 1
B. `-1`
C. 0
D. 4

Answer

To solve the equation \( x^2 + y^2 + 6x + 5 = 4(x - y) \), we will follow these steps:

### Step 1: Rearrange the equation
Start by moving all terms to one side of the equation:
\[ x^2 + y^2 + 6x + 5 - 4x + 4y = 0 \]
This simplifies to:
\[ x^2 + y^2 + 2x + 4y + 5 = 0 \]

### Step 2: Group the terms
Next, we will group the \( x \) and \( y \) terms:
\[ x^2 + 2x + y^2 + 4y + 5 = 0 \]

### Step 3: Complete the square for \( x \)
To complete the square for the \( x \) terms:
\[ x^2 + 2x = (x + 1)^2 - 1 \]
So we rewrite the equation:
\[ (x + 1)^2 - 1 + y^2 + 4y + 5 = 0 \]

### Step 4: Complete the square for \( y \)
Now, complete the square for the \( y \) terms:
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 + 5 = 0 \]
This simplifies to:
\[ (x + 1)^2 + (y + 2)^2 = 0 \]

### Step 5: Analyze the equation
Since the sum of squares equals zero, both squares must individually equal zero:
\[ (x + 1)^2 = 0 \quad \text{and} \quad (y + 2)^2 = 0 \]

### Step 6: Solve for \( x \) and \( y \)
From \( (x + 1)^2 = 0 \):
\[ x + 1 = 0 \implies x = -1 \]
From \( (y + 2)^2 = 0 \):
\[ y + 2 = 0 \implies y = -2 \]

### Step 7: Find \( x - y \)
Now, we need to find \( x - y \):
\[ x - y = -1 - (-2) = -1 + 2 = 1 \]

### Final Answer
Thus, the value of \( x - y \) is \( 1 \).