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

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

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Answer 1 · 2015-11-13T19:17:36

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

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

Explanation:

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

$y^{2} = 16 - x^{2}$ $y^2 = 16-x^2$

Then taking square roots of both sides:

$y = \pm \sqrt{16 - x^{2}} = \pm \sqrt{4^{2} - x^{2}}$ $y = +-sqrt(16-x^2) = +-sqrt(4^2-x^2)$

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

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

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

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