Find the Domain (x+6)^2+(y-5)^2=36

${(x + 6)}^{2} + {(y - 5)}^{2} = 36$

Step 1

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

${(y - 5)}^{2} = 36 - {(x + 6)}^{2}$

Step 2

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

$y - 5 = \pm \sqrt{36 - {(x + 6)}^{2}}$

Step 3

Simplify $\pm \sqrt{36 - {(x + 6)}^{2}}$ .

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

$y - 5 = \pm \sqrt{6^{2} - {(x + 6)}^{2}}$

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

$y - 5 = \pm \sqrt{(6 + x + 6) (6 - (x + 6))}$

Simplify.

Add $6$ and $6$ .

$y - 5 = \pm \sqrt{(x + 12) (6 - (x + 6))}$

Apply the distributive property.

$y - 5 = \pm \sqrt{(x + 12) (6 - x - 1 \cdot 6)}$

Multiply $- 1$ by $6$ .

$y - 5 = \pm \sqrt{(x + 12) (6 - x - 6)}$

Subtract $6$ from $6$ .

$y - 5 = \pm \sqrt{(x + 12) (- x + 0)}$

Add $- x$ and $0$ .

$y - 5 = \pm \sqrt{(x + 12) \cdot - 1 x}$

Factor out negative.

$y - 5 = \pm \sqrt{- ((x + 12) x)}$

$y - 5 = \pm \sqrt{- (x + 12) x}$

Reorder factors in $\pm \sqrt{- (x + 12) x}$ .

$y - 5 = \pm \sqrt{- x (x + 12)}$

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.

$y - 5 = \sqrt{- x (x + 12)}$

Add $5$ to both sides of the equation.

$y = \sqrt{- x (x + 12)} + 5$

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

$y - 5 = - \sqrt{- x (x + 12)}$

Add $5$ to both sides of the equation.

$y = - \sqrt{- x (x + 12)} + 5$

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

$y = \sqrt{- x (x + 12)} + 5$

$y = - \sqrt{- x (x + 12)} + 5$

$y = \sqrt{- x (x + 12)} + 5$

$y = - \sqrt{- x (x + 12)} + 5$

Step 5

Set the radicand in $\sqrt{- x (x + 12)}$ greater than or equal to $0$ to find where the expression is defined.

$- x (x + 12) \geq 0$

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 $0$ .

$x = 0$

$x + 12 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 12$ equal to $0$ and solve for $x$ .

Set $x + 12$ equal to $0$ .

$x + 12 = 0$

Subtract $12$ from both sides of the equation.

$x = - 12$

The final solution is all the values that make $- x (x + 12) \geq 0$ true.

$x = 0, - 12$

Use each root to create test intervals.

$x < - 12$

$- 12 < x < 0$

$x > 0$

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 $x < - 12$ to see if it makes the inequality true.

Choose a value on the interval $x < - 12$ and see if this value makes the original inequality true.

$x = - 14$

Replace $x$ with $- 14$ in the original inequality.

$14 ((- 14) + 12) \geq 0$

The left side $- 28$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 12 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 12 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$2 ((- 2) + 12) \geq 0$

The left side $20$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 0$ to see if it makes the inequality true.

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$- 2 ((2) + 12) \geq 0$

The left side $- 28$ 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.

$x < - 12$ False

$- 12 < x < 0$ True

$x > 0$ False

$x < - 12$ False

$- 12 < x < 0$ True

$x > 0$ False

The solution consists of all of the true intervals.

$- 12 \leq x \leq 0$

Step 7

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

Interval Notation:

$[- 12, 0]$

Set-Builder Notation:

${x | - 12 \leq x \leq 0}$

Step 8

Find the Domain (x+6)^2+(y-5)^2=36

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