Find the Domain (x+2)^2+(y-5)^2=9

${(x + 2)}^{2} + {(y - 5)}^{2} = 9$

Step 1

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

${(y - 5)}^{2} = 9 - {(x + 2)}^{2}$

Step 2

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

$y - 5 = \pm \sqrt{9 - {(x + 2)}^{2}}$

Step 3

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

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

$y - 5 = \pm \sqrt{3^{2} - {(x + 2)}^{2}}$

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 $b = x + 2$ .

$y - 5 = \pm \sqrt{(3 + x + 2) (3 - (x + 2))}$

Simplify.

Add $3$ and $2$ .

$y - 5 = \pm \sqrt{(x + 5) (3 - (x + 2))}$

Apply the distributive property.

$y - 5 = \pm \sqrt{(x + 5) (3 - x - 1 \cdot 2)}$

Multiply $- 1$ by $2$ .

$y - 5 = \pm \sqrt{(x + 5) (3 - x - 2)}$

Subtract $2$ from $3$ .

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

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.

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

Add $5$ to both sides of the equation.

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

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

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

Add $5$ to both sides of the equation.

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

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

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

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

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

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

Step 5

Set the radicand in $\sqrt{(x + 5) (- x + 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 5) (- x + 1) \geq 0$

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 $0$ .

$x + 5 = 0$

$- x + 1 = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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 $(x + 5) (- x + 1) \geq 0$ true.

$x = - 5, 1$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$((- 8) + 5) (- (- 8) + 1) \geq 0$

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 $- 5 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 5) (- (0) + 1) \geq 0$

The left side $5$ 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.

$((4) + 5) (- (4) + 1) \geq 0$

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.

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- 5 \leq x \leq 1$

Step 7

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

Interval Notation:

$[- 5, 1]$

Set-Builder Notation:

${x | - 5 \leq x \leq 1}$

Step 8

Find the Domain (x+2)^2+(y-5)^2=9

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