Domain of `sqrt(a^(2)-x^(2))` `(agt0)` is

Options

A. `[-a,a]`
B. ` (-a,a)`
C. ` [0,a]`
D. ` [-a,0]`

Explanation

To find the domain of the function \( f(x) = \sqrt{a^2 - x^2} \) where \( a > 0 \), we need to ensure that the expression inside the square root is non-negative. This is because the square root function is only defined for non-negative values in the real number system.

### Step-by-Step Solution:

1. **Set up the inequality**:
We need to find when \( a^2 - x^2 \geq 0 \).

2. **Rearrange the inequality**:
This can be rearranged to:
\[ a^2 \geq x^2 \]

3. **Factor the expression**:
We can factor this inequality:
\[ a^2 - x^2 = (a - x)(a + x) \geq 0 \]

4. **Determine the critical points**:
The critical points occur when \( a - x = 0 \) or \( a + x = 0 \). Solving these gives:
- \( x = a \)
- \( x = -a \)

5. **Analyze the intervals**:
We will analyze the sign of the expression \( (a - x)(a + x) \) in the intervals determined by the critical points \( -a \) and \( a \):
- For \( x < -a \): Both \( (a - x) \) and \( (a + x) \) are positive, so the product is positive.
- For \( -a < x < a \): \( (a - x) \) is positive and \( (a + x) \) is negative, so the product is negative.
- For \( x > a \): Both \( (a - x) \) and \( (a + x) \) are negative, so the product is positive.

6. **Determine the valid interval**:
We need the product \( (a - x)(a + x) \) to be non-negative. This occurs in the intervals:
- \( x \leq -a \) or \( x \geq a \)
However, we also need to include the endpoints \( x = -a \) and \( x = a \) since the inequality is non-strict (greater than or equal to).

7. **Conclusion**:
Therefore, the domain of \( f(x) = \sqrt{a^2 - x^2} \) is:
\[ x \in [-a, a] \]

### Final Answer:
The domain of \( \sqrt{a^2 - x^2} \) where \( a > 0 \) is \( [-a, a] \).

---