Question

Does the equation `x^2+y^2=1` describe a function?

Answer 1

The equation `x^2+y^2=1` describes a relation, but not a function.

Explanation:

• A relation between `x` and `y` is a set of pairs `(x, y)`.

• An equation in variables `x, y` always describes a relation - namely the set of pairs of values `(x, y)` which satisfy the equation.

• A function is a relation that has at most one pair `(x, y)` for any value of `x`.

So in our example, the quadratic equation:

`x^2+y^2 = 1`

describes a relation between `x` and `y`.

We can tell whether it defines a function by testing whether given any value of `x` there is at most one value `y` for which `(x, y)` satisfies the equation.

Try `x=0`:

`color(blue)(0)^2+y^2 = 1`

That is `y^2=1`, which has two solutions:

`y = 1" "` or `" "y = -1`

So both of the points `(0, 1)` and `(0, -1)` are in the relation described by the equation.

So it is not a function.

• Graphically `x^2+y^2 = 1` describes the unit circle.

• The vertical line test asks whether any vertical line will intersect the graph in at most one point.

• The line `x=0` intersects the circle in two points.

• So `x^2+y^2=1` fails the vertical line test.

graph{(x^2+y^2-1)(x+0.0001y)(x^2+(y-1)^2-0.004)(x^2+(y+1)^2-0.004)=0 [-4.2, 4.2, -2.1, 2.1]}

More generally, given:

`x^2+y^2=1`

we can attempt to solve for `y` in terms of `x`...

Subtract `x^2` from both sides to get:

`y^2 = 1-x^2`

Take the square root of both sides, allowing for both square roots, to get:

`y = +-sqrt(1-x^2)`

We see that this does not define a unique value of `y` for each value of `x`. That is, it does not describe a function.