Question

How do you decide whether the relation `(x-3)^2 + (y+1)^2 = 25` defines a function?

Answer 1

This is a circle of radius `5` centred at `(3, -1)`.

It fails the vertical line test.

For example, the line `x = 3` intersects the circle at `(3, 4)` and `(3, -6)`

Explanation:

For this equation to define `y` as a function of `x`, there can be at most one value of `y` for each value of `x`.

This is expressed in the vertical line test. Any vertical line may only intersect the relation at at most one point in order that the relation be considered a function.

graph{((x-3)^2+(y+1)^2-25)(x-3+y*0.0001) = 0 [-20, 20, -10, 10]}