Graph x^2+y^2>=36 x^2+9(y+6)^2<=36

$x^{2} + y^{2} \geq 36$ $x^{2} + 9 {(y + 6)}^{2} \leq 36$

Step 1

Solve for $y$ .

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

$y^{2} \geq 36 - x^{2}$

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

$\sqrt{y^{2}} \geq \sqrt{36 - x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \geq \sqrt{36 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{36 - x^{2}}$ .

Rewrite $36$ as $6^{2}$ .

$| y | \geq \sqrt{6^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6$ and $b = x$ .

$| y | \geq \sqrt{(6 + x) (6 - x)}$

Write $| y | \geq \sqrt{(6 + x) (6 - x)}$ 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.

$y \geq \sqrt{(6 + x) (6 - x)}$

Find the domain of $y \geq \sqrt{(6 + x) (6 - x)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \geq \sqrt{(6 + x) (6 - x)}$ .

Set the radicand in $\sqrt{(6 + x) (6 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(6 + x) (6 - x) \geq 0$

Solve for $y$ .

Simplify $(6 + x) (6 - x)$ .

Expand $(6 + x) (6 - x)$ using the FOIL Method.

Apply the distributive property.

$6 (6 - x) + x (6 - x) \geq 0$

Apply the distributive property.

$6 \cdot 6 + 6 (- x) + x (6 - x) \geq 0$

Apply the distributive property.

$6 \cdot 6 + 6 (- x) + x \cdot 6 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $6$ by $6$ .

$36 + 6 (- x) + x \cdot 6 + x (- x) \geq 0$

Multiply $- 1$ by $6$ .

$36 - 6 x + x \cdot 6 + x (- x) \geq 0$

Move $6$ to the left of $x$ .

$36 - 6 x + 6 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$36 - 6 x + 6 x - x \cdot x \geq 0$

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

Move $x$ .

$36 - 6 x + 6 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$36 - 6 x + 6 x - x^{2} \geq 0$

Add $- 6 x$ and $6 x$ .

$36 + 0 - x^{2} \geq 0$

Add $36$ and $0$ .

$36 - x^{2} \geq 0$

Subtract $36$ from both sides of the inequality.

$- x^{2} \geq - 36$

Divide each term in $- x^{2} \geq - 36$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 36$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 36}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 36}{- 1}$

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

$x^{2} \leq \frac{- 36}{- 1}$

Simplify the right side.

Divide $- 36$ by $- 1$ .

$x^{2} \leq 36$

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

$\sqrt{x^{2}} \leq \sqrt{36}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{36}$

Simplify the right side.

Simplify $\sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$| x | \leq \sqrt{6^{2}}$

Pull terms out from under the radical.

$| x | \leq | 6 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$| x | \leq 6$

Write $| x | \leq 6$ 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.

$x \leq 6$

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 $- 1$ .

$- x \leq 6$

Write as a piecewise.

${\begin{cases} x \leq 6 & x \geq 0 \\ - x \leq 6 & x < 0 \end{cases}$

Find the intersection of $x \leq 6$ and $x \geq 0$ .

$0 \leq x \leq 6$

Solve $- x \leq 6$ when $x < 0$ .

Divide each term in $- x \leq 6$ by $- 1$ and simplify.

Divide each term in $- x \leq 6$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{6}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{6}{- 1}$

Simplify the right side.

Divide $6$ by $- 1$ .

$x \geq - 6$

Find the intersection of $x \geq - 6$ and $x < 0$ .

$- 6 \leq x < 0$

Find the union of the solutions.

$- 6 \leq x \leq 6$

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

$[- 6, 6]$

Find the intersection of $y \geq 0$ and $[- 6, 6]$ .

$0 \leq y \leq 6$

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 $- 1$ .

$- y \geq \sqrt{(6 + x) (6 - x)}$

Find the domain of $- y \geq \sqrt{(6 + x) (6 - x)}$ and find the intersection with $y < 0$ .

Find the domain of $- y \geq \sqrt{(6 + x) (6 - x)}$ .

Set the radicand in $\sqrt{(6 + x) (6 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(6 + x) (6 - x) \geq 0$

Solve for $y$ .

Simplify $(6 + x) (6 - x)$ .

Expand $(6 + x) (6 - x)$ using the FOIL Method.

Apply the distributive property.

$6 (6 - x) + x (6 - x) \geq 0$

Apply the distributive property.

$6 \cdot 6 + 6 (- x) + x (6 - x) \geq 0$

Apply the distributive property.

$6 \cdot 6 + 6 (- x) + x \cdot 6 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $6$ by $6$ .

$36 + 6 (- x) + x \cdot 6 + x (- x) \geq 0$

Multiply $- 1$ by $6$ .

$36 - 6 x + x \cdot 6 + x (- x) \geq 0$

Move $6$ to the left of $x$ .

$36 - 6 x + 6 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$36 - 6 x + 6 x - x \cdot x \geq 0$

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

Move $x$ .

$36 - 6 x + 6 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$36 - 6 x + 6 x - x^{2} \geq 0$

Add $- 6 x$ and $6 x$ .

$36 + 0 - x^{2} \geq 0$

Add $36$ and $0$ .

$36 - x^{2} \geq 0$

Subtract $36$ from both sides of the inequality.

$- x^{2} \geq - 36$

Divide each term in $- x^{2} \geq - 36$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 36$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 36}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 36}{- 1}$

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

$x^{2} \leq \frac{- 36}{- 1}$

Simplify the right side.

Divide $- 36$ by $- 1$ .

$x^{2} \leq 36$

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

$\sqrt{x^{2}} \leq \sqrt{36}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{36}$

Simplify the right side.

Simplify $\sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$| x | \leq \sqrt{6^{2}}$

Pull terms out from under the radical.

$| x | \leq | 6 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$| x | \leq 6$

Write $| x | \leq 6$ 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.

$x \leq 6$

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 $- 1$ .

$- x \leq 6$

Write as a piecewise.

${\begin{cases} x \leq 6 & x \geq 0 \\ - x \leq 6 & x < 0 \end{cases}$

Find the intersection of $x \leq 6$ and $x \geq 0$ .

$0 \leq x \leq 6$

Solve $- x \leq 6$ when $x < 0$ .

Divide each term in $- x \leq 6$ by $- 1$ and simplify.

Divide each term in $- x \leq 6$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{6}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{6}{- 1}$

Simplify the right side.

Divide $6$ by $- 1$ .

$x \geq - 6$

Find the intersection of $x \geq - 6$ and $x < 0$ .

$- 6 \leq x < 0$

Find the union of the solutions.

$- 6 \leq x \leq 6$

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

$[- 6, 6]$

Find the intersection of $y < 0$ and $[- 6, 6]$ .

$- 6 \leq y < 0$

Write as a piecewise.

${\begin{cases} y \geq \sqrt{(6 + x) (6 - x)} & 0 \leq y \leq 6 \\ - y \geq \sqrt{(6 + x) (6 - x)} & - 6 \leq y < 0 \end{cases}$

Find the intersection of $y \geq \sqrt{(6 + x) (6 - x)}$ and $0 \leq y \leq 6$ .

No solution

Solve $- y \geq \sqrt{(6 + x) (6 - x)}$ when $- 6 \leq y < 0$ .

Divide each term in $- y \geq \sqrt{(6 + x) (6 - x)}$ by $- 1$ and simplify.

Divide each term in $- y \geq \sqrt{(6 + x) (6 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{\sqrt{(6 + x) (6 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{(6 + x) (6 - x)}}{- 1}$

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{(6 + x) (6 - x)}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{(6 + x) (6 - x)}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{(6 + x) (6 - x)}$

Rewrite $- 1 \cdot \sqrt{(6 + x) (6 - x)}$ as $- \sqrt{(6 + x) (6 - x)}$ .

$y \leq - \sqrt{(6 + x) (6 - x)}$

Find the intersection of $y \leq - \sqrt{(6 + x) (6 - x)}$ and $- 6 \leq y < 0$ .

$- 6 \leq y < N o (Min i m u m)$

Find the union of the solutions.

$- 6 \leq y < i N o Min m^{2} u$

Step 2

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

$9 {(y + 6)}^{2} \leq 36 - x^{2}$

Step 3

Plot each graph on the same coordinate system.

$x^{2} + y^{2} \geq 36$

$x^{2} + 9 {(y + 6)}^{2} \leq 36$

Step 4

Graph x^2+y^2>=36 x^2+9(y+6)^2<=36

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