Question

If \[ R=\{(x,y)\mid x,y\in \mathbb{R},\ x^2+y^2=1\} \] is a relation in \(\mathbb{R}\), then \(R\) is: 

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Equivalence Relation

Hint:

For relation-property questions:
• Reflexive \(\Rightarrow\) check whether \((x,x)\) belongs to the relation for all elements.
• Symmetric \(\Rightarrow\) interchange \(x\) and \(y\) and verify whether the condition remains unchanged.
• Transitive \(\Rightarrow\) usually test with a counterexample.
• Equivalence relation requires all three properties simultaneously. Relations involving expressions like \(x^2+y^2\) are often symmetric because swapping variables does not change the equation.

Answer 1

The Correct Option is B

Concept: A relation \(R\) on a set \(A\) may possess different properties such as reflexivity, symmetry, and transitivity. The important definitions are:
• Reflexive Relation: A relation \(R\) is reflexive if: \[ (x,x)\in R \quad \text{for every } x\in A \]
• Symmetric Relation: A relation \(R\) is symmetric if: \[ (x,y)\in R \Rightarrow (y,x)\in R \]
• Transitive Relation: A relation \(R\) is transitive if: \[ (x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R \]
• Equivalence Relation: A relation is called an equivalence relation if it is simultaneously: \[ \text{reflexive, symmetric, and transitive} \] We now check each property carefully for the given relation.

Step 1: Understanding the given relation The relation is: \[ R=\{(x,y)\mid x,y\in \mathbb{R},\ x^2+y^2=1\} \] This means an ordered pair \((x,y)\) belongs to \(R\) only when: \[ x^2+y^2=1 \] Geometrically, this represents all points lying on the unit circle centered at the origin.

Step 2: Checking whether the relation is reflexive For reflexivity, we must have: \[ (x,x)\in R \quad \text{for every } x\in \mathbb{R} \] Substitute \(y=x\) into the relation: \[ x^2+x^2=1 \] \[ 2x^2=1 \] \[ x^2=\frac12 \] \[ x=\pm \frac{1}{\sqrt2} \] Thus, only two specific real numbers satisfy the condition. But reflexivity requires the condition to hold for every real number. For example, take \(x=0\): \[ 0^2+0^2=0\neq 1 \] Hence, \[ (0,0)\notin R \] Therefore, the relation is not reflexive.

Step 3: Checking whether the relation is symmetric Suppose: \[ (x,y)\in R \] Then by definition, \[ x^2+y^2=1 \] Now interchange \(x\) and \(y\): \[ y^2+x^2=1 \] Since addition is commutative, \[ y^2+x^2=x^2+y^2 \] Therefore, \[ (y,x)\in R \] Hence, whenever \((x,y)\in R\), we also have \((y,x)\in R\). Therefore, the relation is symmetric.

Step 4: Checking whether the relation is transitive For transitivity, we require: \[ (x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R \] We test this using a counterexample. Take: \[ (x,y)=\left(1,0\right) \] Then, \[ 1^2+0^2=1 \] Hence, \[ (1,0)\in R \] Now take: \[ (y,z)=\left(0,1\right) \] Then, \[ 0^2+1^2=1 \] Thus, \[ (0,1)\in R \] Now check whether: \[ (1,1)\in R \] We compute: \[ 1^2+1^2=2 \] Since: \[ 2\neq 1 \] we get: \[ (1,1)\notin R \] Thus, the transitive condition fails. Therefore, the relation is not transitive.

Step 5: Checking whether it is an equivalence relation An equivalence relation must be: \[ \text{reflexive + symmetric + transitive} \] But this relation is only symmetric and is neither reflexive nor transitive. Hence, it is not an equivalence relation. Final Conclusion: The given relation satisfies only the symmetric property. Therefore, the correct answer is: \[ \boxed{(B)\ \text{Symmetric}} \]