If the equation `x^2-y^2-2x-2y+c=0` represent an empty set , then find the value of c.

Answer

To solve the problem, we need to determine the value of \( c \) such that the equation \( x^2 - y^2 - 2x - 2y + c = 0 \) represents an empty set. This means that the equation has no real solutions.

### Step-by-Step Solution:

1. **Rearrange the Equation**:
We start with the given equation:
\[ x^2 - y^2 - 2x - 2y + c = 0 \]
We can rearrange it to express it as a quadratic equation in terms of \( x \):
\[ x^2 - 2x + (c - y^2 - 2y) = 0 \]

2. **Identify Coefficients**:
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = -2 \)
- \( c = c - y^2 - 2y \)

3. **Condition for No Real Solutions**:
For the quadratic equation to have no real solutions, the discriminant \( D \) must be less than 0:
\[ D = b^2 - 4ac < 0 \]
Substituting the values we identified:
\[ D = (-2)^2 - 4(1)(c - y^2 - 2y) < 0 \]
This simplifies to:
\[ 4 - 4(c - y^2 - 2y) < 0 \]

4. **Simplify the Inequality**:
Rearranging the inequality gives:
\[ 4 - 4c + 4y^2 + 8y < 0 \]
Dividing the entire inequality by 4:
\[ 1 - c + y^2 + 2y < 0 \]

5. **Rearranging for \( c \)**:
We can rearrange the inequality to isolate \( c \):
\[ c > 1 + y^2 + 2y \]

6. **Finding the Minimum Value of \( y^2 + 2y \)**:
The expression \( y^2 + 2y \) can be rewritten as:
\[ y^2 + 2y = (y + 1)^2 - 1 \]
The minimum value of \( (y + 1)^2 \) is 0 (when \( y = -1 \)), thus:
\[ y^2 + 2y \geq -1 \]
Therefore:
\[ 1 + y^2 + 2y \geq 1 - 1 = 0 \]

7. **Conclusion**:
Since \( c > 1 + y^2 + 2y \) must hold for all \( y \), and the minimum value of \( 1 + y^2 + 2y \) is 0, we conclude:
\[ c > 0 \]
However, for the equation to represent an empty set, we need \( c \) to be less than or equal to 0. Thus, the only value that satisfies this condition is:
\[ c < 0 \]

### Final Answer:
The value of \( c \) must be less than 0:
\[ c < 0 \]