If `R={(x,y),x, y, in W,x^(2)+y^(2)le 4` then domain of R is

Options

A. `{0,1,2,3}
B. {0,1,2}
C. {0,1}
D. {0,2,3}

Answer

To find the domain of the relation \( R = \{(x, y) | x, y \in W, x^2 + y^2 \leq 4\} \), where \( W \) is the set of whole numbers, we will follow these steps:

### Step 1: Understand the condition
The condition \( x^2 + y^2 \leq 4 \) represents a circle with a radius of 2 centered at the origin in the Cartesian plane. Since \( x \) and \( y \) must be whole numbers, we need to find all pairs \( (x, y) \) that satisfy this condition.

**Hint:** Remember that whole numbers are non-negative integers (0, 1, 2, ...).

### Step 2: Set values for \( x \)
We will consider different values for \( x \) starting from 0 and moving upwards, as \( x \) must be a whole number.

1. **For \( x = 0 \)**:
\[ 0^2 + y^2 \leq 4 \implies y^2 \leq 4 \implies y = 0, 1, 2 \]
Possible pairs: \( (0, 0), (0, 1), (0, 2) \)

2. **For \( x = 1 \)**:
\[ 1^2 + y^2 \leq 4 \implies 1 + y^2 \leq 4 \implies y^2 \leq 3 \implies y = 0, 1 \]
Possible pairs: \( (1, 0), (1, 1) \)

3. **For \( x = 2 \)**:
\[ 2^2 + y^2 \leq 4 \implies 4 + y^2 \leq 4 \implies y^2 \leq 0 \implies y = 0 \]
Possible pair: \( (2, 0) \)

### Step 3: Collect all unique \( x \) values
From the pairs we found, the possible values of \( x \) are:
- From \( x = 0 \): 0
- From \( x = 1 \): 1
- From \( x = 2 \): 2

### Step 4: Compile the domain
The domain of the relation \( R \) consists of all unique \( x \) values we found:
\[ \text{Domain of } R = \{0, 1, 2\} \]

### Final Answer:
The domain of \( R \) is \( \{0, 1, 2\} \).

---