Find the Domain x^2+(y-6)^2=4

$x^{2} + {(y - 6)}^{2} = 4$

Step 1

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

${(y - 6)}^{2} = 4 - x^{2}$

Step 2

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

$y - 6 = \pm \sqrt{4 - x^{2}}$

Step 3

Simplify $\pm \sqrt{4 - x^{2}}$ .

Rewrite $4$ as $2^{2}$ .

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

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

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

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.

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

Add $6$ to both sides of the equation.

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

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

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

Add $6$ to both sides of the equation.

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

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

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

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

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

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

Step 5

Set the radicand in $\sqrt{(2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

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

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 $0$ .

$2 + x = 0$

$2 - x = 0$

Set $2 + x$ equal to $0$ and solve for $x$ .

Set $2 + x$ equal to $0$ .

$2 + x = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $2 - x$ equal to $0$ and solve for $x$ .

Set $2 - x$ equal to $0$ .

$2 - x = 0$

Solve $2 - x = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

The final solution is all the values that make $(2 + x) (2 - x) \geq 0$ true.

$x = - 2, 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$x > 2$

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 $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

$(2 - 4) (2 - (- 4)) \geq 0$

The left side $- 12$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$(2 + 4) (2 - (4)) \geq 0$

The left side $- 12$ 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.

$x < - 2$ False

$- 2 < x < 2$ True

$x > 2$ False

$x < - 2$ False

$- 2 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 2 \leq x \leq 2$

Step 7

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

Interval Notation:

$[- 2, 2]$

Set-Builder Notation:

${x | - 2 \leq x \leq 2}$

Step 8

Find the Domain x^2+(y-6)^2=4

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