Find the Domain (x+3)^2+y^2=36

${(x + 3)}^{2} + y^{2} = 36$

Step 1

Subtract ${(x + 3)}^{2}$ from both sides of the equation.

$y^{2} = 36 - {(x + 3)}^{2}$

Step 2

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

$y = \pm \sqrt{36 - {(x + 3)}^{2}}$

Step 3

Simplify $\pm \sqrt{36 - {(x + 3)}^{2}}$ .

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

$y = \pm \sqrt{6^{2} - {(x + 3)}^{2}}$

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 $b = x + 3$ .

$y = \pm \sqrt{(6 + x + 3) (6 - (x + 3))}$

Simplify.

Add $6$ and $3$ .

$y = \pm \sqrt{(x + 9) (6 - (x + 3))}$

Apply the distributive property.

$y = \pm \sqrt{(x + 9) (6 - x - 1 \cdot 3)}$

Multiply $- 1$ by $3$ .

$y = \pm \sqrt{(x + 9) (6 - x - 3)}$

Subtract $3$ from $6$ .

$y = \pm \sqrt{(x + 9) (- x + 3)}$

Step 4

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.

$y = \sqrt{(x + 9) (- x + 3)}$

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

$y = - \sqrt{(x + 9) (- x + 3)}$

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

$y = \sqrt{(x + 9) (- x + 3)}$

$y = - \sqrt{(x + 9) (- x + 3)}$

$y = \sqrt{(x + 9) (- x + 3)}$

$y = - \sqrt{(x + 9) (- x + 3)}$

Step 5

Set the radicand in $\sqrt{(x + 9) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 6

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 9 = 0$

$- x + 3 = 0$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

Set $- x + 3$ equal to $0$ and solve for $x$ .

Set $- x + 3$ equal to $0$ .

$- x + 3 = 0$

Solve $- x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

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

Divide each term in $- x = - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

The final solution is all the values that make $(x + 9) (- x + 3) \geq 0$ true.

$x = - 9, 3$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$((- 12) + 9) (- (- 12) + 3) \geq 0$

The left side $- 45$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 9 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

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

The left side $27$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$((6) + 9) (- (6) + 3) \geq 0$

The left side $- 45$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 9$ False

$- 9 < x < 3$ True

$x > 3$ False

$x < - 9$ False

$- 9 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 9 \leq x \leq 3$

Step 7

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

Interval Notation:

$[- 9, 3]$

Set-Builder Notation:

${x | - 9 \leq x \leq 3}$

Step 8

Find the Domain (x+3)^2+y^2=36

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