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Given (x, y) ∈ R, where x² + y² = 16, we know y = ±√(16 − x²). When x = 0, y equals ±4. For x = ±4, y equals 0. It is observed that no other integer pairs (x, y) satisfy x² + y² = 16. The set R is defined as {(0, 4), (0, −4), (4, 0), (−4, 0)}. What is one value in the domain of R?
- x = 0
- x = 4
- x = −4
- x = ±4
Correct answer: x = 4
Solution
The equation x² + y² = 16 represents a circle with radius 4. When x = 4, substituting into the equation gives y = 0, which is one of the integer solutions.
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