Find the Domain and Range (x+2)^2+y^2=25

${(x + 2)}^{2} + y^{2} = 25$

Step 1

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

$y^{2} = 25 - {(x + 2)}^{2}$

Step 2

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

$y = \pm \sqrt{25 - {(x + 2)}^{2}}$

Step 3

Simplify $\pm \sqrt{25 - {(x + 2)}^{2}}$ .

Rewrite $25$ as $5^{2}$ .

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

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

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

Simplify.

Add $5$ and $2$ .

$y = \pm \sqrt{(x + 7) (5 - (x + 2))}$

Apply the distributive property.

$y = \pm \sqrt{(x + 7) (5 - x - 1 \cdot 2)}$

Multiply $- 1$ by $2$ .

$y = \pm \sqrt{(x + 7) (5 - x - 2)}$

Subtract $2$ from $5$ .

$y = \pm \sqrt{(x + 7) (- x + 3)}$

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.

$y = \sqrt{(x + 7) (- x + 3)}$

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

$y = - \sqrt{(x + 7) (- x + 3)}$

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

$y = \sqrt{(x + 7) (- x + 3)}$

$y = - \sqrt{(x + 7) (- x + 3)}$

$y = \sqrt{(x + 7) (- x + 3)}$

$y = - \sqrt{(x + 7) (- x + 3)}$

Step 5

Set the radicand in $\sqrt{(x + 7) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 7) (- x + 3) \geq 0$

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 $0$ .

$x + 7 = 0$

$- x + 3 = 0$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

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 $(x + 7) (- x + 3) \geq 0$ true.

$x = - 7, 3$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 3$

$x > 3$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

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

$((- 10) + 7) (- (- 10) + 3) \geq 0$

The left side $- 39$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 7 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 7) (- (0) + 3) \geq 0$

The left side $21$ 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.

$((6) + 7) (- (6) + 3) \geq 0$

The left side $- 39$ 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.

$x < - 7$ False

$- 7 < x < 3$ True

$x > 3$ False

$x < - 7$ False

$- 7 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 7 \leq x \leq 3$

Step 7

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

Interval Notation:

$[- 7, 3]$

Set-Builder Notation:

${x | - 7 \leq x \leq 3}$

Step 8

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${y | - 5 \leq y \leq 5}$

Step 9

Determine the domain and range.

Domain: $[- 7, 3], {x | - 7 \leq x \leq 3}$

Range: $[- 5, 5], {y | - 5 \leq y \leq 5}$

Step 10

Find the Domain and Range (x+2)^2+y^2=25

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