Graph (x-3)^2+y^2<=9

${(x - 3)}^{2} + y^{2} \leq 9$

Step 1

Subtract ${(x - 3)}^{2}$ from both sides of the inequality.

$y^{2} \leq 9 - {(x - 3)}^{2}$

Step 2

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{y^{2}} \leq \sqrt{9 - {(x - 3)}^{2}}$

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{9 - {(x - 3)}^{2}}$

Simplify the right side.

Simplify $\sqrt{9 - {(x - 3)}^{2}}$ .

Rewrite $9$ as $3^{2}$ .

$| y | \leq \sqrt{3^{2} - {(x - 3)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x - 3$ .

$| y | \leq \sqrt{(3 + x - 3) (3 - (x - 3))}$

Simplify.

Subtract $3$ from $3$ .

$| y | \leq \sqrt{(x + 0) (3 - (x - 3))}$

Add $x$ and $0$ .

$| y | \leq \sqrt{x (3 - (x - 3))}$

Apply the distributive property.

$| y | \leq \sqrt{x (3 - x - - 3)}$

Multiply $- 1$ by $- 3$ .

$| y | \leq \sqrt{x (3 - x + 3)}$

Add $3$ and $3$ .

$| y | \leq \sqrt{x (- x + 6)}$

Step 4

Write $| y | \leq \sqrt{x (- x + 6)}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

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

Find the domain of $y \leq \sqrt{x (- x + 6)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{x (- x + 6)}$ .

Set the radicand in $\sqrt{x (- x + 6)}$ greater than or equal to $0$ to find where the expression is defined.

$x (- x + 6) \geq 0$

Solve for $y$ .

Simplify $x (- x + 6)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (- x) + x \cdot 6 \geq 0$

Reorder.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 6 \geq 0$

Move $6$ to the left of $x$ .

$- x \cdot x + 6 \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- (x \cdot x) + 6 \cdot x \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + 6 \cdot x \geq 0$

$- x^{2} + 6 x \geq 0$

Convert the inequality to an equation.

$- x^{2} + 6 x = 0$

Factor $- x$ out of $- x^{2} + 6 x$ .

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 6 x = 0$

Factor $- x$ out of $6 x$ .

$- x \cdot x - x \cdot - 6 = 0$

Factor $- x$ out of $- x (x) - x (- 6)$ .

$- x (x - 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $- x (x - 6) = 0$ true.

$x = 0, 6$

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

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 $x < 0$ 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.

$(- 2) (- (- 2) + 6) \geq 0$

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

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

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$(8) (- (8) + 6) \geq 0$

The left side $- 16$ 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.

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$0 \leq x \leq 6$

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

$[0, 6]$

Find the intersection of $y \geq 0$ and $[0, 6]$ .

$0 \leq y \leq 6$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

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

Find the domain of $- y \leq \sqrt{x (- x + 6)}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \sqrt{x (- x + 6)}$ .

Set the radicand in $\sqrt{x (- x + 6)}$ greater than or equal to $0$ to find where the expression is defined.

$x (- x + 6) \geq 0$

Solve for $y$ .

Simplify $x (- x + 6)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (- x) + x \cdot 6 \geq 0$

Reorder.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 6 \geq 0$

Move $6$ to the left of $x$ .

$- x \cdot x + 6 \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- (x \cdot x) + 6 \cdot x \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + 6 \cdot x \geq 0$

$- x^{2} + 6 x \geq 0$

Convert the inequality to an equation.

$- x^{2} + 6 x = 0$

Factor $- x$ out of $- x^{2} + 6 x$ .

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 6 x = 0$

Factor $- x$ out of $6 x$ .

$- x \cdot x - x \cdot - 6 = 0$

Factor $- x$ out of $- x (x) - x (- 6)$ .

$- x (x - 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $- x (x - 6) = 0$ true.

$x = 0, 6$

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

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 $x < 0$ 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.

$(- 2) (- (- 2) + 6) \geq 0$

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

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

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$(8) (- (8) + 6) \geq 0$

The left side $- 16$ 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.

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < 0$ False

$0 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$0 \leq x \leq 6$

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

$[0, 6]$

Find the intersection of $y < 0$ and $[0, 6]$ .

$No solution$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{x (- x + 6)} & 0 \leq y \leq 6 \end{cases}$

Step 5

Find the intersection of $y \leq \sqrt{x (- x + 6)}$ and $0 \leq y \leq 6$ .

$0 \leq y \leq N o (Min i m u m)$

Step 6

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Step 7

Graph (x-3)^2+y^2<=9

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