Find the Domain and Range x^2+y^2=16

Rewrite ${\displaystyle 16}$ as ${\displaystyle 4^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=4}$ and ${\displaystyle b=x}$.

First, use the positive value of the ${\displaystyle \pm }$ to find the first solution.

Next, use the negative value of the ${\displaystyle \pm }$ to find the second solution.

The complete solution is the result of both the positive and negative portions of the solution.

If any individual factor on the left side of the equation is equal to ${\displaystyle 0}$, the entire expression will be equal to ${\displaystyle 0}$.

Set ${\displaystyle 4+x}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

Set ${\displaystyle 4-x}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

The final solution is all the values that make ${\displaystyle (4+x)(4-x)\ge 0}$ true.

Use each root to create test intervals.

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

${\displaystyle x<-4}$ False

${\displaystyle -4<x<4}$ True

${\displaystyle x>4}$ False

The solution consists of all of the true intervals.