Given `R={(x,y):x,y in W, x^(2)+y^(2) =169}`, then the domain of R is

Options

A. {0,5, 12, 13}
B. {-13,-12,-5, 0,5, 12, 13}
C. {0, 1, 2, 3, ... 13}
D. `{0, +-1, +-2,...+-13}`

Answer

To find the domain of the relation \( R = \{(x,y) : x,y \in W, x^2 + y^2 = 169\} \), where \( W \) represents the set of whole numbers (non-negative integers), we will follow these steps:

### Step 1: Understand the equation
The equation \( x^2 + y^2 = 169 \) represents a circle with a radius of 13 centered at the origin in the coordinate plane. We need to find all pairs of whole numbers \( (x, y) \) that satisfy this equation.

### Step 2: Express \( y \) in terms of \( x \)
From the equation, we can express \( y \) as:
\[ y = \sqrt{169 - x^2} \]
Since \( y \) must be a whole number, \( 169 - x^2 \) must be a perfect square.

### Step 3: Determine the possible values for \( x \)
Since \( x \) is a whole number, we need to find values of \( x \) such that \( 169 - x^2 \geq 0 \). This means:
\[ x^2 \leq 169 \implies -13 \leq x \leq 13 \]
However, since \( x \) must be a whole number, we restrict \( x \) to the range \( 0 \leq x \leq 13 \).

### Step 4: Check each value of \( x \)
We will check each whole number value of \( x \) from 0 to 13 to see if \( y \) is also a whole number.

- For \( x = 0 \):
\[ y = \sqrt{169 - 0^2} = \sqrt{169} = 13 \quad (\text{whole number}) \]

- For \( x = 1 \):
\[ y = \sqrt{169 - 1^2} = \sqrt{168} \quad (\text{not a whole number}) \]

- For \( x = 2 \):
\[ y = \sqrt{169 - 2^2} = \sqrt{165} \quad (\text{not a whole number}) \]

- For \( x = 3 \):
\[ y = \sqrt{169 - 3^2} = \sqrt{160} \quad (\text{not a whole number}) \]

- For \( x = 4 \):
\[ y = \sqrt{169 - 4^2} = \sqrt{153} \quad (\text{not a whole number}) \]

- For \( x = 5 \):
\[ y = \sqrt{169 - 5^2} = \sqrt{144} = 12 \quad (\text{whole number}) \]

- For \( x = 6 \):
\[ y = \sqrt{169 - 6^2} = \sqrt{133} \quad (\text{not a whole number}) \]

- For \( x = 7 \):
\[ y = \sqrt{169 - 7^2} = \sqrt{120} \quad (\text{not a whole number}) \]

- For \( x = 8 \):
\[ y = \sqrt{169 - 8^2} = \sqrt{105} \quad (\text{not a whole number}) \]

- For \( x = 9 \):
\[ y = \sqrt{169 - 9^2} = \sqrt{88} \quad (\text{not a whole number}) \]

- For \( x = 10 \):
\[ y = \sqrt{169 - 10^2} = \sqrt{69} \quad (\text{not a whole number}) \]

- For \( x = 11 \):
\[ y = \sqrt{169 - 11^2} = \sqrt{48} \quad (\text{not a whole number}) \]

- For \( x = 12 \):
\[ y = \sqrt{169 - 12^2} = \sqrt{25} = 5 \quad (\text{whole number}) \]

- For \( x = 13 \):
\[ y = \sqrt{169 - 13^2} = \sqrt{0} = 0 \quad (\text{whole number}) \]

### Step 5: Collect valid pairs
The valid pairs \( (x,y) \) where both \( x \) and \( y \) are whole numbers are:
- \( (0, 13) \)
- \( (5, 12) \)
- \( (12, 5) \)
- \( (13, 0) \)

### Step 6: Identify the domain
The domain of the relation \( R \) consists of the \( x \)-values from the valid pairs:
- \( 0, 5, 12, 13 \)

Thus, the domain of \( R \) is:
\[ \text{Domain of } R = \{0, 5, 12, 13\} \]