Find the Domain (x+1)^2+(y-2)^2=16

${(x + 1)}^{2} + {(y - 2)}^{2} = 16$

Step 1

Subtract ${(x + 1)}^{2}$ from both sides of the equation.

${(y - 2)}^{2} = 16 - {(x + 1)}^{2}$

Step 2

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

$y - 2 = \pm \sqrt{16 - {(x + 1)}^{2}}$

Step 3

Simplify $\pm \sqrt{16 - {(x + 1)}^{2}}$ .

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

$y - 2 = \pm \sqrt{4^{2} - {(x + 1)}^{2}}$

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 $b = x + 1$ .

$y - 2 = \pm \sqrt{(4 + x + 1) (4 - (x + 1))}$

Simplify.

Add $4$ and $1$ .

$y - 2 = \pm \sqrt{(x + 5) (4 - (x + 1))}$

Apply the distributive property.

$y - 2 = \pm \sqrt{(x + 5) (4 - x - 1 \cdot 1)}$

Multiply $- 1$ by $1$ .

$y - 2 = \pm \sqrt{(x + 5) (4 - x - 1)}$

Subtract $1$ from $4$ .

$y - 2 = \pm \sqrt{(x + 5) (- x + 3)}$

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.

$y - 2 = \sqrt{(x + 5) (- x + 3)}$

Add $2$ to both sides of the equation.

$y = \sqrt{(x + 5) (- x + 3)} + 2$

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

$y - 2 = - \sqrt{(x + 5) (- x + 3)}$

Add $2$ to both sides of the equation.

$y = - \sqrt{(x + 5) (- x + 3)} + 2$

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

$y = \sqrt{(x + 5) (- x + 3)} + 2$

$y = - \sqrt{(x + 5) (- x + 3)} + 2$

$y = \sqrt{(x + 5) (- x + 3)} + 2$

$y = - \sqrt{(x + 5) (- x + 3)} + 2$

Step 5

Set the radicand in $\sqrt{(x + 5) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 5) (- x + 3) \geq 0$

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 $0$ .

$x + 5 = 0$

$- x + 3 = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

$- x + 3 = 0$

Solve $- x + 3 = 0$ 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 $(x + 5) (- x + 3) \geq 0$ true.

$x = - 5, 3$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$x > 3$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

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

$((- 8) + 5) (- (- 8) + 3) \geq 0$

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 $- 5 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 5) (- (0) + 3) \geq 0$

The left side $15$ 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.

$((6) + 5) (- (6) + 3) \geq 0$

The left side $- 33$ 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.

$x < - 5$ False

$- 5 < x < 3$ True

$x > 3$ False

$x < - 5$ False

$- 5 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 5 \leq x \leq 3$

Step 7

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

Interval Notation:

$[- 5, 3]$

Set-Builder Notation:

${x | - 5 \leq x \leq 3}$

Step 8

Find the Domain (x+1)^2+(y-2)^2=16

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