The number of pairs (x,y) which will satisfy the equation `x^2-x y+y^2=4(x+y-4)` is

Options

A. 1
B. 2
C. 4
D. none of these

Explanation

To solve the equation \( x^2 - xy + y^2 = 4(x + y - 4) \) and find the number of pairs \((x, y)\) that satisfy it, we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ x^2 - xy + y^2 = 4(x + y - 4) \]
Expanding the right side:
\[ x^2 - xy + y^2 = 4x + 4y - 16 \]
Now, rearranging all terms to one side gives:
\[ x^2 - xy + y^2 - 4x - 4y + 16 = 0 \]

### Step 2: Grouping Terms
We can rewrite the equation as:
\[ x^2 - 4x - xy + y^2 - 4y + 16 = 0 \]

### Step 3: Completing the Square
Next, we will complete the square for the \(x\) and \(y\) terms.

For the \(x\) terms:
\[ x^2 - 4x = (x - 2)^2 - 4 \]

For the \(y\) terms:
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 - xy + (y - 2)^2 - 4 + 16 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 2)^2 - xy + 8 = 0 \]

### Step 4: Rearranging Again
Now we rearrange it to isolate the terms:
\[ (x - 2)^2 + (y - 2)^2 - xy = -8 \]

### Step 5: Analyzing the Equation
The expression \((x - 2)^2 + (y - 2)^2\) is always non-negative, and for the equation to hold, the left side must equal \(-8\), which is impossible since a sum of squares cannot be negative.

### Conclusion
The only way for the equation to hold true is if all terms are zero:
1. \(x - y = 0 \Rightarrow x = y\)
2. \(x - 4 = 0 \Rightarrow x = 4\)
3. \(y - 4 = 0 \Rightarrow y = 4\)

Thus, the only solution is:
\[ (x, y) = (4, 4) \]

### Final Answer
The number of pairs \((x, y)\) that satisfy the equation is:
\[ \text{1 pair: } (4, 4) \]