Graph x^2+y^2>16 - Math solver

$x^{2} + y^{2} > 16$

Subtract $x^{2}$ from both sides of the inequality.

$y^{2} > 16 - x^{2}$

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

$y > \pm \sqrt{16 - x^{2}}$

Simplify the right side of the equation.

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

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

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

$y > \pm \sqrt{(4 + x) (4 - x)}$

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

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

$y > \sqrt{(4 + x) (4 - x)}$

Next, use the negative value of the $\pm$ to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the $-$ portion of the solution.

$y < - \sqrt{(4 + x) (4 - x)}$

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

$y > \sqrt{(4 + x) (4 - x)}$ or $y < - \sqrt{(4 + x) (4 - x)}$

Graph x^2+y^2>16

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