Find the Domain (x+4)^2+(y-3)^2=16

${(x + 4)}^{2} + {(y - 3)}^{2} = 16$

Step 1

Subtract ${(x + 4)}^{2}$ from both sides of the equation.

${(y - 3)}^{2} = 16 - {(x + 4)}^{2}$

Step 2

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

$y - 3 = \pm \sqrt{16 - {(x + 4)}^{2}}$

Step 3

Simplify $\pm \sqrt{16 - {(x + 4)}^{2}}$ .

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

$y - 3 = \pm \sqrt{4^{2} - {(x + 4)}^{2}}$

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 $b = x + 4$ .

$y - 3 = \pm \sqrt{(4 + x + 4) (4 - (x + 4))}$

Simplify.

Add $4$ and $4$ .

$y - 3 = \pm \sqrt{(x + 8) (4 - (x + 4))}$

Apply the distributive property.

$y - 3 = \pm \sqrt{(x + 8) (4 - x - 1 \cdot 4)}$

Multiply $- 1$ by $4$ .

$y - 3 = \pm \sqrt{(x + 8) (4 - x - 4)}$

Subtract $4$ from $4$ .

$y - 3 = \pm \sqrt{(x + 8) (- x + 0)}$

Add $- x$ and $0$ .

$y - 3 = \pm \sqrt{(x + 8) \cdot - 1 x}$

Factor out negative.

$y - 3 = \pm \sqrt{- ((x + 8) x)}$

$y - 3 = \pm \sqrt{- (x + 8) x}$

Reorder factors in $\pm \sqrt{- (x + 8) x}$ .

$y - 3 = \pm \sqrt{- x (x + 8)}$

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.

$y - 3 = \sqrt{- x (x + 8)}$

Add $3$ to both sides of the equation.

$y = \sqrt{- x (x + 8)} + 3$

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

$y - 3 = - \sqrt{- x (x + 8)}$

Add $3$ to both sides of the equation.

$y = - \sqrt{- x (x + 8)} + 3$

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

$y = \sqrt{- x (x + 8)} + 3$

$y = - \sqrt{- x (x + 8)} + 3$

$y = \sqrt{- x (x + 8)} + 3$

$y = - \sqrt{- x (x + 8)} + 3$

Step 5

Set the radicand in $\sqrt{- x (x + 8)}$ greater than or equal to $0$ to find where the expression is defined.

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

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 $0$ .

$x = 0$

$x + 8 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $- x (x + 8) \geq 0$ true.

$x = 0, - 8$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 0$

$x > 0$

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$10 ((- 10) + 8) \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < 0$ and see if this value makes the original inequality true.

$x = - 4$

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

$4 ((- 4) + 8) \geq 0$

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 $x > 0$ to see if it makes the inequality true.

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$- 2 ((2) + 8) \geq 0$

The left side $- 20$ 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.

$x < - 8$ False

$- 8 < x < 0$ True

$x > 0$ False

$x < - 8$ False

$- 8 < x < 0$ True

$x > 0$ False

The solution consists of all of the true intervals.

$- 8 \leq x \leq 0$

Step 7

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

Interval Notation:

$[- 8, 0]$

Set-Builder Notation:

${x | - 8 \leq x \leq 0}$

Step 8

Find the Domain (x+4)^2+(y-3)^2=16

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