Which of the following relations are odd?
I. y=2
II. `y=x`
III. `x^(2)+y^(2)=1`

Options

A. only II
B. only I and III
C. only II and III
D. I,II and III

Explanation

To determine which of the given relations are odd, we need to check each relation against the definition of an odd function. A function \( f(x) \) is considered odd if it satisfies the condition:

\[ f(-x) = -f(x) \]

for all \( x \) in the domain of \( f \). Let's analyze each of the given relations one by one.

### Step 1: Analyze the first relation \( y = 2 \)

1. **Replace \( y \) with \( f(x) \)**:
\[ f(x) = 2 \]

2. **Calculate \( f(-x) \)**:
\[ f(-x) = 2 \]

3. **Calculate \( -f(x) \)**:
\[ -f(x) = -2 \]

4. **Check the odd function condition**:
\[ f(-x) \neq -f(x) \quad \text{(since } 2 \neq -2\text{)} \]

**Conclusion**: The first relation \( y = 2 \) is **not an odd function**.

### Step 2: Analyze the second relation \( y = x \)

1. **Replace \( y \) with \( f(x) \)**:
\[ f(x) = x \]

2. **Calculate \( f(-x) \)**:
\[ f(-x) = -x \]

3. **Calculate \( -f(x) \)**:
\[ -f(x) = -x \]

4. **Check the odd function condition**:
\[ f(-x) = -f(x) \quad \text{(since } -x = -x\text{)} \]

**Conclusion**: The second relation \( y = x \) is an **odd function**.

### Step 3: Analyze the third relation \( x^2 + y^2 = 1 \)

1. **This is not a function but an equation of a circle**. To check if it is odd, we will replace \( x \) with \( -x \) and \( y \) with \( -y \).

2. **Replace \( x \) and \( y \)**:
\[ (-x)^2 + (-y)^2 = 1 \]

3. **Simplify**:
\[ x^2 + y^2 = 1 \]

4. **Check if the equation remains the same**:
\[ x^2 + y^2 = 1 \quad \text{(is the same as the original equation)} \]

**Conclusion**: The third relation \( x^2 + y^2 = 1 \) is **symmetric about the origin**, hence it is an **odd relation**.

### Final Conclusion

- **First relation**: Not odd
- **Second relation**: Odd
- **Third relation**: Odd

Thus, the odd relations are **II and III**.

### Summary of the Answer
The odd relations are **II and III**.