Find the Domain (x+5)^2+(y-1)^2=36

${(x + 5)}^{2} + {(y - 1)}^{2} = 36$

Step 1

Subtract ${(x + 5)}^{2}$ from both sides of the equation.

${(y - 1)}^{2} = 36 - {(x + 5)}^{2}$

Step 2

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

$y - 1 = \pm \sqrt{36 - {(x + 5)}^{2}}$

Step 3

Simplify $\pm \sqrt{36 - {(x + 5)}^{2}}$ .

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

$y - 1 = \pm \sqrt{6^{2} - {(x + 5)}^{2}}$

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 $b = x + 5$ .

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

Simplify.

Add $6$ and $5$ .

$y - 1 = \pm \sqrt{(x + 11) (6 - (x + 5))}$

Apply the distributive property.

$y - 1 = \pm \sqrt{(x + 11) (6 - x - 1 \cdot 5)}$

Multiply $- 1$ by $5$ .

$y - 1 = \pm \sqrt{(x + 11) (6 - x - 5)}$

Subtract $5$ from $6$ .

$y - 1 = \pm \sqrt{(x + 11) (- x + 1)}$

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.

$y - 1 = \sqrt{(x + 11) (- x + 1)}$

Add $1$ to both sides of the equation.

$y = \sqrt{(x + 11) (- x + 1)} + 1$

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

$y - 1 = - \sqrt{(x + 11) (- x + 1)}$

Add $1$ to both sides of the equation.

$y = - \sqrt{(x + 11) (- x + 1)} + 1$

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

$y = \sqrt{(x + 11) (- x + 1)} + 1$

$y = - \sqrt{(x + 11) (- x + 1)} + 1$

$y = \sqrt{(x + 11) (- x + 1)} + 1$

$y = - \sqrt{(x + 11) (- x + 1)} + 1$

Step 5

Set the radicand in $\sqrt{(x + 11) (- x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 11) (- x + 1) \geq 0$

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 $0$ .

$x + 11 = 0$

$- x + 1 = 0$

Set $x + 11$ equal to $0$ and solve for $x$ .

Set $x + 11$ equal to $0$ .

$x + 11 = 0$

Subtract $11$ from both sides of the equation.

$x = - 11$

Set $- x + 1$ equal to $0$ and solve for $x$ .

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $(x + 11) (- x + 1) \geq 0$ true.

$x = - 11, 1$

Use each root to create test intervals.

$x < - 11$

$- 11 < x < 1$

$x > 1$

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 $x < - 11$ to see if it makes the inequality true.

Choose a value on the interval $x < - 11$ and see if this value makes the original inequality true.

$x = - 14$

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

$((- 14) + 11) (- (- 14) + 1) \geq 0$

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 $- 11 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 11 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 11) (- (0) + 1) \geq 0$

The left side $11$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$((4) + 11) (- (4) + 1) \geq 0$

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.

$x < - 11$ False

$- 11 < x < 1$ True

$x > 1$ False

$x < - 11$ False

$- 11 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- 11 \leq x \leq 1$

Step 7

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

Interval Notation:

$[- 11, 1]$

Set-Builder Notation:

${x | - 11 \leq x \leq 1}$

Step 8

Find the Domain (x+5)^2+(y-1)^2=36

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