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Arnav Colon

Answered question

2022-12-25

How to decide whether the relation

x^{2} + y^{2} = 16

defines a function?

tjaran12f

Beginner2022-12-26Added 13 answers

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 = \pm \sqrt{16 - x^{2}} = \pm \sqrt{4^{2} - x^{2}}

Now

\sqrt{4^{2} - x^{2}}

only takes Real values when

4^{2} - x^{2} \geq 0

, that is when

- 4 \leq x \leq 4

. For values of

x \in (- 4, 4)

, there are two values of

y

. The vertical line test indicates that this relation is not a function.
graph{x^2+y^2=16 [-10, 10, -5, 5]}

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