Find the Domain (x+6)^2+(y-1)^2=9

${(x + 6)}^{2} + {(y - 1)}^{2} = 9$

Step 1

Subtract ${(x + 6)}^{2}$ from both sides of the equation.

${(y - 1)}^{2} = 9 - {(x + 6)}^{2}$

Step 2

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

$y - 1 = \pm \sqrt{9 - {(x + 6)}^{2}}$

Step 3

Simplify $\pm \sqrt{9 - {(x + 6)}^{2}}$ .

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

$y - 1 = \pm \sqrt{3^{2} - {(x + 6)}^{2}}$

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 $b = x + 6$ .

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

Simplify.

Add $3$ and $6$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $6$ .

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

Subtract $6$ from $3$ .

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

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.

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

Add $1$ to both sides of the equation.

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

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

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

Add $1$ to both sides of the equation.

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

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

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

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

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

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

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 $x$ .

Add $3$ to 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 $- 27$ 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.

$x = - 6$

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

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

The left side $9$ 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.

$x = 0$

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

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

The left side $- 27$ 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+6)^2+(y-1)^2=9

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