Graph x^2+y^2<9

$x^{2} + y^{2} < 9$

Step 1

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

$y^{2} < 9 - x^{2}$

Step 2

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

$\sqrt{y^{2}} < \sqrt{9 - x^{2}}$

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | < \sqrt{9 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{9 - x^{2}}$ .

Rewrite $9$ as $3^{2}$ .

$| y | < \sqrt{3^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x$ .

$| y | < \sqrt{(3 + x) (3 - x)}$

Step 4

Write $| y | < \sqrt{(3 + x) (3 - x)}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

$y < \sqrt{(3 + x) (3 - x)}$

Find the domain of $y < \sqrt{(3 + x) (3 - x)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y < \sqrt{(3 + x) (3 - x)}$ .

Set the radicand in $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

Solve for $y$ .

Simplify $(3 + x) (3 - x)$ .

Expand $(3 + x) (3 - x)$ using the FOIL Method.

Apply the distributive property.

$3 (3 - x) + x (3 - x) \geq 0$

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x (3 - x) \geq 0$

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Multiply $- 1$ by $3$ .

$9 - 3 x + x \cdot 3 + x (- x) \geq 0$

Move $3$ to the left of $x$ .

$9 - 3 x + 3 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$9 - 3 x + 3 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$9 - 3 x + 3 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$9 - 3 x + 3 x - x^{2} \geq 0$

Add $- 3 x$ and $3 x$ .

$9 + 0 - x^{2} \geq 0$

Add $9$ and $0$ .

$9 - x^{2} \geq 0$

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq - 9$

Divide each term in $- x^{2} \geq - 9$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 9}{- 1}$

Simplify the right side.

Divide $- 9$ by $- 1$ .

$x^{2} \leq 9$

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

$\sqrt{x^{2}} \leq \sqrt{9}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{9}$

Simplify the right side.

Simplify $\sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$| x | \leq \sqrt{3^{2}}$

Pull terms out from under the radical.

$| x | \leq | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq 3$

Write $| x | \leq 3$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 3$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 3$

Write as a piecewise.

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Find the intersection of $x \leq 3$ and $x \geq 0$ .

$0 \leq x \leq 3$

Solve $- x \leq 3$ when $x < 0$ .

Divide each term in $- x \leq 3$ by $- 1$ and simplify.

Divide each term in $- x \leq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{3}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x \geq - 3$

Find the intersection of $x \geq - 3$ and $x < 0$ .

$- 3 \leq x < 0$

Find the union of the solutions.

$- 3 \leq x \leq 3$

The domain is all values of $y$ that make the expression defined.

$[- 3, 3]$

Find the intersection of $y \geq 0$ and $[- 3, 3]$ .

$0 \leq y \leq 3$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y < \sqrt{(3 + x) (3 - x)}$

Find the domain of $- y < \sqrt{(3 + x) (3 - x)}$ and find the intersection with $y < 0$ .

Find the domain of $- y < \sqrt{(3 + x) (3 - x)}$ .

Set the radicand in $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

Solve for $y$ .

Simplify $(3 + x) (3 - x)$ .

Expand $(3 + x) (3 - x)$ using the FOIL Method.

Apply the distributive property.

$3 (3 - x) + x (3 - x) \geq 0$

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x (3 - x) \geq 0$

Apply the distributive property.

$3 \cdot 3 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 3 (- x) + x \cdot 3 + x (- x) \geq 0$

Multiply $- 1$ by $3$ .

$9 - 3 x + x \cdot 3 + x (- x) \geq 0$

Move $3$ to the left of $x$ .

$9 - 3 x + 3 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$9 - 3 x + 3 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$9 - 3 x + 3 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$9 - 3 x + 3 x - x^{2} \geq 0$

Add $- 3 x$ and $3 x$ .

$9 + 0 - x^{2} \geq 0$

Add $9$ and $0$ .

$9 - x^{2} \geq 0$

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq - 9$

Divide each term in $- x^{2} \geq - 9$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 9}{- 1}$

Simplify the right side.

Divide $- 9$ by $- 1$ .

$x^{2} \leq 9$

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

$\sqrt{x^{2}} \leq \sqrt{9}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{9}$

Simplify the right side.

Simplify $\sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$| x | \leq \sqrt{3^{2}}$

Pull terms out from under the radical.

$| x | \leq | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq 3$

Write $| x | \leq 3$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 3$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 3$

Write as a piecewise.

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Find the intersection of $x \leq 3$ and $x \geq 0$ .

$0 \leq x \leq 3$

Solve $- x \leq 3$ when $x < 0$ .

Divide each term in $- x \leq 3$ by $- 1$ and simplify.

Divide each term in $- x \leq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{3}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x \geq - 3$

Find the intersection of $x \geq - 3$ and $x < 0$ .

$- 3 \leq x < 0$

Find the union of the solutions.

$- 3 \leq x \leq 3$

The domain is all values of $y$ that make the expression defined.

$[- 3, 3]$

Find the intersection of $y < 0$ and $[- 3, 3]$ .

$- 3 \leq y < 0$

Write as a piecewise.

${\begin{cases} y < \sqrt{(3 + x) (3 - x)} & 0 \leq y \leq 3 \\ - y < \sqrt{(3 + x) (3 - x)} & - 3 \leq y < 0 \end{cases}$

Step 5

Find the intersection of $y < \sqrt{(3 + x) (3 - x)}$ and $0 \leq y \leq 3$ .

$0 \leq y \leq N o (Min i m u m)$

Step 6

Solve $- y < \sqrt{(3 + x) (3 - x)}$ when $- 3 \leq y < 0$ .

Divide each term in $- y < \sqrt{(3 + x) (3 - x)}$ by $- 1$ and simplify.

Divide each term in $- y < \sqrt{(3 + x) (3 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Divide $y$ by $1$ .

$y > \frac{\sqrt{(3 + x) (3 - x)}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{(3 + x) (3 - x)}}{- 1}$ .

$y > - 1 \cdot \sqrt{(3 + x) (3 - x)}$

Rewrite $- 1 \cdot \sqrt{(3 + x) (3 - x)}$ as $- \sqrt{(3 + x) (3 - x)}$ .

$y > - \sqrt{(3 + x) (3 - x)}$

Find the intersection of $y > - \sqrt{(3 + x) (3 - x)}$ and $- 3 \leq y < 0$ .

No solution

Step 7

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Step 8

Graph x^2+y^2<9

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