Find the Domain x^2+(y-3)^2=36

$x^{2} + {(y - 3)}^{2} = 36$

Step 1

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

${(y - 3)}^{2} = 36 - x^{2}$

Step 2

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

$y - 3 = \pm \sqrt{36 - x^{2}}$

Step 3

Simplify $\pm \sqrt{36 - x^{2}}$ .

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

$y - 3 = \pm \sqrt{6^{2} - x^{2}}$

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

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

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.

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

Add $3$ to both sides of the equation.

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

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

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

Add $3$ to both sides of the equation.

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

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

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

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

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

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

Step 5

Set the radicand in $\sqrt{(6 + x) (6 - x)}$ greater than or equal to $0$ to find where the expression is defined.

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

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 $0$ .

$6 + x = 0$

$6 - x = 0$

Set $6 + x$ equal to $0$ and solve for $x$ .

Set $6 + x$ equal to $0$ .

$6 + x = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Set $6 - x$ equal to $0$ and solve for $x$ .

Set $6 - x$ equal to $0$ .

$6 - x = 0$

Solve $6 - x = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

The final solution is all the values that make $(6 + x) (6 - x) \geq 0$ true.

$x = - 6, 6$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 6$

$x > 6$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$(6 - 8) (6 - (- 8)) \geq 0$

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 $- 6 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $36$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$(6 + 8) (6 - (8)) \geq 0$

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.

$x < - 6$ False

$- 6 < x < 6$ True

$x > 6$ False

$x < - 6$ False

$- 6 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$- 6 \leq x \leq 6$

Step 7

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

Interval Notation:

$[- 6, 6]$

Set-Builder Notation:

${x | - 6 \leq x \leq 6}$

Step 8

Find the Domain x^2+(y-3)^2=36

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