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

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

Step 1

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

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

Step 2

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

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

Step 3

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

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

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

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 $b = x$ .

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

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.

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

Add $5$ to both sides of the equation.

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

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

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

Add $5$ to both sides of the equation.

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

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

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

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

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

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

Step 5

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

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

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 $0$ .

$3 + x = 0$

$3 - x = 0$

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

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

$3 + x = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$3 - x = 0$

Solve $3 - x = 0$ 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 $(3 + x) (3 - x) \geq 0$ true.

$x = - 3, 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$x > 3$

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 $x < - 3$ 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.

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

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 $- 3 < x < 3$ to see if it makes the inequality true.

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

$x = 0$

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

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

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.

$x = 6$

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

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

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.

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 3$

Step 7

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

Interval Notation:

$[- 3, 3]$

Set-Builder Notation:

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

Step 8

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

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