Question

Given circle P centered at the origin, with a radius of 4 units

Circle P is not a function
domain: -4 ≤ x ≤ 4
f(0) = 0
0 3
1 7
2 11
3 15
Circle P is a function
x²+y²= 16
a. y = x²+3
b. y = sqrt 16-x²
c. y = 4-x²
d. y = sqrt 16 + x²

Answer 1

Yes, the equation (x² + y² = 16) represents Circle P. The correct representation of the equation of Circle P as a function is Option B:

`\(y = √(16 - x^2)\)`

Step-by-step explanation:

Let's evaluate each option for the given circle P centered at the origin with a radius of 4 units and the specified points.

Given circle P:
`\(x^2 + y^2 = 16\)`

a.
`\(y = x^2 + 3\)`

- Substitute (x = 0):
`\(y = 0^2 + 3 = 3\)` (Doesn't satisfy the point (0, 0))

- Substitute (x = 1):
`\(y = 1^2 + 3 = 4\)` (Doesn't satisfy the point (1, 7))

- Incorrect

b.
`\(y = √(16 - x^2)\)`

- Substitute (x = 0):
`\(y = √(16 - 0^2) = 4\)` (Satisfies the point (0, 0))

- Substitute (x = 1):
`\(y = √(16 - 1^2) = √(15)\)`(Doesn't satisfy the point (1, 7))

- Incorrect

c.
`\(y = 4 - x^2\)`

- Substitute (x = 0):
`\(y = 4 - 0^2 = 4\)` (Doesn't satisfy the point (0, 0))

- Substitute (x = 1):
`\(y = 4 - 1^2 = 3\)` (Satisfies the point (1, 7))

- Incorrect

d.
`\(y = √(16 + x^2)\)`

- Substitute (x = 0):
`\(y = √(16 + 0^2) = 4\)` (Satisfies the point (0, 0))

- Substitute (x = 1):
`\(y = √(16 + 1^2) = √(17)\)` (Doesn't satisfy the point (1, 7))

- Incorrect

Therefore, the correct option is B
`\(y = √(16 - x^2)\)`, as it satisfies the given conditions at (x = 0) and does not satisfy the conditions at (x = 1).