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

Oct 31, 2015

This is a circle of radius $5$ centred at $\left(3 , - 1\right)$.

It fails the vertical line test.

For example, the line $x = 3$ intersects the circle at $\left(3 , 4\right)$ and $\left(3 , - 6\right)$

#### 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]}