Evaluate the Function x^2+y^2=4

$x^{2} + y^{2} = 4$

Subtract $y^{2}$ from both sides of the equation.

$x^{2} = 4 - y^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4 - y^{2}}$

Simplify $\pm \sqrt{4 - y^{2}}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2} - y^{2}}$

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

$x = \pm \sqrt{(2 + y) (2 - y)}$

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

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

$x = \sqrt{(2 + y) (2 - y)}$

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

$x = - \sqrt{(2 + y) (2 - y)}$

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

$x = \sqrt{(2 + y) (2 - y)}$

$x = - \sqrt{(2 + y) (2 - y)}$

$x = \sqrt{(2 + y) (2 - y)}$

$x = - \sqrt{(2 + y) (2 - y)}$