Solve for x r = square root of x^2+y^2

$r = \sqrt{x^{2} + y^{2}}$

Rewrite the equation as $\sqrt{x^{2} + y^{2}} = r$ .

$\sqrt{x^{2} + y^{2}} = r$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x^{2} + y^{2}}}^{2} = r^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + y^{2}}$ as ${(x^{2} + y^{2})}^{\frac{1}{2}}$ .

${({(x^{2} + y^{2})}^{\frac{1}{2}})}^{2} = r^{2}$

Simplify the left side.

Simplify ${({(x^{2} + y^{2})}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x^{2} + y^{2})}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x^{2} + y^{2})}^{\frac{1}{2} \cdot 2} = r^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} + y^{2})}^{\frac{1}{2} \cdot 2} = r^{2}$

Rewrite the expression.

${(x^{2} + y^{2})}^{1} = r^{2}$

Simplify.

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

Solve for $x$ .

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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