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

1 Answer
George C.
Nov 13, 2015

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]}

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