Graph (x-1)^2+(y+2)^2<=4

${(x - 1)}^{2} + {(y + 2)}^{2} \leq 4$

Step 1

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

${(y + 2)}^{2} \leq 4 - {(x - 1)}^{2}$

Step 2

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

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

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y + 2 | \leq \sqrt{4 - {(x - 1)}^{2}}$

Simplify the right side.

Simplify $\sqrt{4 - {(x - 1)}^{2}}$ .

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

$| y + 2 | \leq \sqrt{2^{2} - {(x - 1)}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x - 1$ .

$| y + 2 | \leq \sqrt{(2 + x - 1) (2 - (x - 1))}$

Simplify.

Subtract $1$ from $2$ .

$| y + 2 | \leq \sqrt{(x + 1) (2 - (x - 1))}$

Apply the distributive property.

$| y + 2 | \leq \sqrt{(x + 1) (2 - x - - 1)}$

Multiply $- 1$ by $- 1$ .

$| y + 2 | \leq \sqrt{(x + 1) (2 - x + 1)}$

Add $2$ and $1$ .

$| y + 2 | \leq \sqrt{(x + 1) (- x + 3)}$

Step 4

Write $| y + 2 | \leq \sqrt{(x + 1) (- x + 3)}$ as a piecewise.

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

$y + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$y \geq - 2$

In the piece where $y + 2$ is non-negative, remove the absolute value.

$y + 2 \leq \sqrt{(x + 1) (- x + 3)}$

Find the domain of $y + 2 \leq \sqrt{(x + 1) (- x + 3)}$ and find the intersection with $y \geq - 2$ .

Find the domain of $y + 2 \leq \sqrt{(x + 1) (- x + 3)}$ .

Set the radicand in $\sqrt{(x + 1) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (- x + 3) \geq 0$

Solve for $y$ .

Simplify $(x + 1) (- x + 3)$ .

Expand $(x + 1) (- x + 3)$ using the FOIL Method.

Apply the distributive property.

$x (- x + 3) + 1 (- x + 3) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x + 3) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

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

Move $x$ .

$- (x \cdot x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Move $3$ to the left of $x$ .

$- x^{2} + 3 \cdot x + 1 (- x) + 1 \cdot 3 \geq 0$

Multiply $- x$ by $1$ .

$- x^{2} + 3 x - x + 1 \cdot 3 \geq 0$

Multiply $3$ by $1$ .

$- x^{2} + 3 x - x + 3 \geq 0$

Subtract $x$ from $3 x$ .

$- x^{2} + 2 x + 3 \geq 0$

Convert the inequality to an equation.

$- x^{2} + 2 x + 3 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 2 x + 3$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 2 x + 3 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 2 x)$ .

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

Factor.

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$- ((x - 3) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x + 1) = 0$

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

$x - 3 = 0$

$x + 1 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

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 $x < - 1$ 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.

$x = - 4$

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

$((- 4) + 1) (- (- 4) + 3) \geq 0$

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 1) (- (0) + 3) \geq 0$

The left side $3$ 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.

$((6) + 1) (- (6) + 3) \geq 0$

The left side $- 21$ 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.

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 1 \leq x \leq 3$

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

$[- 1, 3]$

Find the intersection of $y \geq - 2$ and $[- 1, 3]$ .

$- 1 \leq y \leq 3$

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

$y + 2 < 0$

Subtract $2$ from both sides of the inequality.

$y < - 2$

In the piece where $y + 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- (y + 2) \leq \sqrt{(x + 1) (- x + 3)}$

Find the domain of $- (y + 2) \leq \sqrt{(x + 1) (- x + 3)}$ and find the intersection with $y < - 2$ .

Find the domain of $- (y + 2) \leq \sqrt{(x + 1) (- x + 3)}$ .

Set the radicand in $\sqrt{(x + 1) (- x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (- x + 3) \geq 0$

Solve for $y$ .

Simplify $(x + 1) (- x + 3)$ .

Expand $(x + 1) (- x + 3)$ using the FOIL Method.

Apply the distributive property.

$x (- x + 3) + 1 (- x + 3) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x + 3) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

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

Move $x$ .

$- (x \cdot x) + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 3 + 1 (- x) + 1 \cdot 3 \geq 0$

Move $3$ to the left of $x$ .

$- x^{2} + 3 \cdot x + 1 (- x) + 1 \cdot 3 \geq 0$

Multiply $- x$ by $1$ .

$- x^{2} + 3 x - x + 1 \cdot 3 \geq 0$

Multiply $3$ by $1$ .

$- x^{2} + 3 x - x + 3 \geq 0$

Subtract $x$ from $3 x$ .

$- x^{2} + 2 x + 3 \geq 0$

Convert the inequality to an equation.

$- x^{2} + 2 x + 3 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 2 x + 3$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 2 x + 3 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 2 x)$ .

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

Factor.

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$- ((x - 3) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x + 1) = 0$

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

$x - 3 = 0$

$x + 1 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

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 $x < - 1$ 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.

$x = - 4$

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

$((- 4) + 1) (- (- 4) + 3) \geq 0$

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 1) (- (0) + 3) \geq 0$

The left side $3$ 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.

$((6) + 1) (- (6) + 3) \geq 0$

The left side $- 21$ 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.

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

$x < - 1$ False

$- 1 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 1 \leq x \leq 3$

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

$[- 1, 3]$

Find the intersection of $y < - 2$ and $[- 1, 3]$ .

$No solution$

Write as a piecewise.

${\begin{cases} y + 2 \leq \sqrt{(x + 1) (- x + 3)} & - 1 \leq y \leq 3 \end{cases}$

Step 5

Solve $y + 2 \leq \sqrt{(x + 1) (- x + 3)}$ when $- 1 \leq y \leq 3$ .

Subtract $2$ from both sides of the inequality.

$y \leq \sqrt{(x + 1) (- x + 3)} - 2$

Find the intersection of $y \leq \sqrt{(x + 1) (- x + 3)} - 2$ and $- 1 \leq y \leq 3$ .

$- 1 \leq y \leq N o (Min i m u m)$

Step 6

Find the union of the solutions.

$- 1 \leq y \leq i N o Min m^{2} u$

Step 7

Graph (x-1)^2+(y+2)^2<=4

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