Find the Domain (x+5)^2+(y-6)^2=16

${(x + 5)}^{2} + {(y - 6)}^{2} = 16$

Step 1

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

${(y - 6)}^{2} = 16 - {(x + 5)}^{2}$

Step 2

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

$y - 6 = \pm \sqrt{16 - {(x + 5)}^{2}}$

Step 3

Simplify $\pm \sqrt{16 - {(x + 5)}^{2}}$ .

Rewrite $16$ as $4^{2}$ .

$y - 6 = \pm \sqrt{4^{2} - {(x + 5)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x + 5$ .

$y - 6 = \pm \sqrt{(4 + x + 5) (4 - (x + 5))}$

Simplify.

Add $4$ and $5$ .

$y - 6 = \pm \sqrt{(x + 9) (4 - (x + 5))}$

Apply the distributive property.

$y - 6 = \pm \sqrt{(x + 9) (4 - x - 1 \cdot 5)}$

Multiply $- 1$ by $5$ .

$y - 6 = \pm \sqrt{(x + 9) (4 - x - 5)}$

Subtract $5$ from $4$ .

$y - 6 = \pm \sqrt{(x + 9) (- x - 1)}$

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.

$y - 6 = \sqrt{(x + 9) (- x - 1)}$

Add $6$ to both sides of the equation.

$y = \sqrt{(x + 9) (- x - 1)} + 6$

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

$y - 6 = - \sqrt{(x + 9) (- x - 1)}$

Add $6$ to both sides of the equation.

$y = - \sqrt{(x + 9) (- x - 1)} + 6$

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

$y = \sqrt{(x + 9) (- x - 1)} + 6$

$y = - \sqrt{(x + 9) (- x - 1)} + 6$

$y = \sqrt{(x + 9) (- x - 1)} + 6$

$y = - \sqrt{(x + 9) (- x - 1)} + 6$

Step 5

Set the radicand in $\sqrt{(x + 9) (- x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 9) (- x - 1) \geq 0$

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 $0$ .

$x + 9 = 0$

$- x - 1 = 0$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

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

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

$- x - 1 = 0$

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

Add $1$ to 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 $(x + 9) (- x - 1) \geq 0$ true.

$x = - 9, - 1$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 1$

$x > - 1$

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 $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$((- 12) + 9) (- (- 12) - 1) \geq 0$

The left side $- 33$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 9 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < - 1$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$((- 5) + 9) (- (- 5) - 1) \geq 0$

The left side $16$ 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.

$x = 0$

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

$((0) + 9) (- (0) - 1) \geq 0$

The left side $- 9$ 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.

$x < - 9$ False

$- 9 < x < - 1$ True

$x > - 1$ False

$x < - 9$ False

$- 9 < x < - 1$ True

$x > - 1$ False

The solution consists of all of the true intervals.

$- 9 \leq x \leq - 1$

Step 7

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

Interval Notation:

$[- 9, - 1]$

Set-Builder Notation:

${x | - 9 \leq x \leq - 1}$

Step 8

Find the Domain (x+5)^2+(y-6)^2=16

Related Questions

Support LCMGCF.com by disabling your adblocker.