Let `P={(x,y)|x^(2)+y^(2)=1,x,yinR}`. Then, R, is

Options

A. Reflexive
B. Symmetric
C. Transitive
D. Anti-symmetric

Explanation

To determine the properties of the relation \( R = \{(x,y) | x^2 + y^2 = 1, x, y \in \mathbb{R}\} \), we will check if it is reflexive, symmetric, transitive, or anti-symmetric.

### Step 1: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( x \) in the set, the pair \( (x,x) \) is in \( R \).

- For \( (x,x) \) to be in \( R \), we need:
\[ x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]
- The pairs \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) and \( \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \) are in \( R \), but not all real numbers \( x \) have the pair \( (x,x) \) in \( R \).

**Conclusion**: The relation \( R \) is **not reflexive**.

### Step 2: Check for Symmetry
A relation \( R \) is symmetric if whenever \( (x,y) \) is in \( R \), then \( (y,x) \) is also in \( R \).

- Assume \( (x,y) \in R \), meaning \( x^2 + y^2 = 1 \).
- If we switch \( x \) and \( y \), we have \( y^2 + x^2 = 1 \), which is the same equation.

**Conclusion**: The relation \( R \) is **symmetric**.

### Step 3: Check for Transitivity
A relation \( R \) is transitive if whenever \( (x,y) \) and \( (y,z) \) are in \( R \), then \( (x,z) \) must also be in \( R \).

- Assume \( (x,y) \in R \) and \( (y,z) \in R \):
\[ x^2 + y^2 = 1 \quad \text{and} \quad y^2 + z^2 = 1 \]
- From these, we cannot conclude that \( x^2 + z^2 = 1 \). For example, let \( x = 1, y = 0, z = 1 \):
- \( (1,0) \in R \) and \( (0,1) \in R \), but \( (1,1) \notin R \) since \( 1^2 + 1^2 \neq 1 \).

**Conclusion**: The relation \( R \) is **not transitive**.

### Step 4: Check for Anti-symmetry
A relation \( R \) is anti-symmetric if whenever \( (x,y) \in R \) and \( (y,x) \in R \), then \( x = y \).

- Since we found that \( R \) is symmetric, it cannot be anti-symmetric unless \( x = y \) for all pairs, which is not the case here.

**Conclusion**: The relation \( R \) is **not anti-symmetric**.

### Final Conclusion
The relation \( R \) is **symmetric** but neither reflexive, transitive, nor anti-symmetric.

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