Let `P(x,y)` be a point which is a constant distance from the origin. Then equivalence relation of `(1,-1)` is

Options

A. `A={(x,y)abs(x^2+y^2=1)}`
B. `A={(x,y)abs(x^2+y^2=2)}`
C. `A={(x,y)abs(x^2+y^2=3)}`
D. `A={(x,y)abs(x^2+y^2=4)}`

Answer

To solve the problem, we need to determine the equivalence relation of the point \( (1, -1) \) given that the point \( P(x, y) \) is at a constant distance from the origin.

### Step-by-step Solution:

1. **Understanding the Distance from the Origin**:
The distance \( d \) of a point \( P(x, y) \) from the origin \( (0, 0) \) is given by the distance formula:
\[ d = \sqrt{x^2 + y^2} \]
If this distance is constant, we can denote it as \( k \). Therefore, we have:
\[ \sqrt{x^2 + y^2} = k \]
Squaring both sides gives:
\[ x^2 + y^2 = k^2 \]

2. **Substituting the Point (1, -1)**:
Now, we need to check the equivalence relation of the point \( (1, -1) \). We will substitute \( x = 1 \) and \( y = -1 \) into the equation:
\[ x^2 + y^2 = 1^2 + (-1)^2 \]
Calculating this gives:
\[ x^2 + y^2 = 1 + 1 = 2 \]

3. **Finding the Value of \( k^2 \)**:
From our earlier equation \( x^2 + y^2 = k^2 \), we can now equate:
\[ k^2 = 2 \]

4. **Conclusion**:
The equivalence relation for the point \( (1, -1) \) is that it lies on the circle with radius \( k = \sqrt{2} \) centered at the origin. Thus, the equation representing this equivalence relation is:
\[ x^2 + y^2 = 2 \]

### Final Answer:
The equivalence relation of the point \( (1, -1) \) is given by the equation:
\[ x^2 + y^2 = 2 \]