A circle touches the lines `x-y- 1 =0 and x -y +1 =0.` the centre of the circle lies on the line

Options

A. `x +y -1 =0`
B. `x + y + 1=0`
C. `x -y =0`
D. none of these

Answer

To solve the problem, we need to find the line on which the center of a circle lies, given that the circle touches the lines \(x - y - 1 = 0\) and \(x - y + 1 = 0\).

### Step-by-Step Solution:

1. **Identify the Given Lines**:
The equations of the lines are:
\[ L_1: x - y - 1 = 0 \quad \text{(or } y = x - 1\text{)} \]
\[ L_2: x - y + 1 = 0 \quad \text{(or } y = x + 1\text{)} \]
These lines are parallel because they have the same slope (1).

2. **Determine the Distance Between the Lines**:
The distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by the formula:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For our lines:
- \(C_1 = -1\) (from \(L_1\))
- \(C_2 = 1\) (from \(L_2\))
- \(A = 1\), \(B = -1\)

Plugging in these values:
\[ d = \frac{|1 - (-1)|}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

3. **Find the Midpoint Between the Lines**:
The center of the circle will be equidistant from both lines, so it will lie on the line that is the average of the two lines. The midpoint \(M\) between the two lines can be found by averaging their y-intercepts:
- The y-intercept of \(L_1\) is \(-1\).
- The y-intercept of \(L_2\) is \(1\).

The average y-intercept is:
\[ y = \frac{-1 + 1}{2} = 0 \]
Therefore, the line that represents the midpoint is:
\[ y = x \]

4. **Conclusion**:
The center of the circle lies on the line:
\[ y = x \]