Question

How do you decide whether the relation `x^2 +y^2 =16` defines a function?

Answer 1

This is the equation of a circle of radius `4` with centre `(0, 0)`.

If you attempt to define `y` in terms of `x` you get multiple values for `x in (-4, 4)`.

Explanation:

Starting with `x^2+y^2=16`, subtract `x^2` from both sides to get:

`y^2 = 16-x^2`

Then taking square roots of both sides:

`y = +-sqrt(16-x^2) = +-sqrt(4^2-x^2)`

Now `sqrt(4^2-x^2)` only takes Real values when `4^2-x^2 >= 0`, that is when `-4 <= x <= 4`. For values of `x in (-4, 4)`, there are two values of `y`. So this relation fails the vertical line test and is not a function.

graph{x^2+y^2=16 [-10, 10, -5, 5]}