Graph x^2+y^2<49 x^2+y^2>25

$x^{2} + y^{2} < 49$ $x^{2} + y^{2} > 25$

Solve for $y$ .

Subtract $x^{2}$ from both sides of the inequality.

$y^{2} < 49 - x^{2}$

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

$\sqrt{y^{2}} < \sqrt{49 - x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | < \sqrt{49 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{49 - x^{2}}$ .

Rewrite $49$ as $7^{2}$ .

$| y | < \sqrt{7^{2} - x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = x$ .

$| y | < \sqrt{(7 + x) (7 - x)}$

Write $| y | < \sqrt{(7 + x) (7 - x)}$ 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.

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

Find the domain of $y < \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y < \sqrt{(7 + x) (7 - x)}$ .

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

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

Solve for $y$ .

Simplify $(7 + x) (7 - x)$ .

Expand $(7 + x) (7 - x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

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

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 49}{- 1}$

Simplify the right side.

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Simplify the right side.

Simplify $\sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Pull terms out from under the radical.

$| x | \leq | 7 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Write $| x | \leq 7$ as a piecewise.

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

$x \geq 0$

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

$x \leq 7$

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

$x < 0$

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

$- x \leq 7$

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Solve $- x \leq 7$ when $x < 0$ .

Divide each term in $- x \leq 7$ by $- 1$ and simplify.

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{7}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x \geq - 7$

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Find the union of the solutions.

$- 7 \leq x \leq 7$

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

$[- 7, 7]$

Find the intersection of $y \geq 0$ and $[- 7, 7]$ .

$0 \leq y \leq 7$

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 $- 1$ .

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

Find the domain of $- y < \sqrt{(7 + x) (7 - x)}$ and find the intersection with $y < 0$ .

Find the domain of $- y < \sqrt{(7 + x) (7 - x)}$ .

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

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

Solve for $y$ .

Simplify $(7 + x) (7 - x)$ .

Expand $(7 + x) (7 - x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $7$ .

$49 + 7 (- x) + x \cdot 7 + x (- x) \geq 0$

Multiply $- 1$ by $7$ .

$49 - 7 x + x \cdot 7 + x (- x) \geq 0$

Move $7$ to the left of $x$ .

$49 - 7 x + 7 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$49 - 7 x + 7 x - x \cdot x \geq 0$

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

Move $x$ .

$49 - 7 x + 7 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$49 - 7 x + 7 x - x^{2} \geq 0$

Add $- 7 x$ and $7 x$ .

$49 + 0 - x^{2} \geq 0$

Add $49$ and $0$ .

$49 - x^{2} \geq 0$

Subtract $49$ from both sides of the inequality.

$- x^{2} \geq - 49$

Divide each term in $- x^{2} \geq - 49$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 49$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 49}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 49}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 49}{- 1}$

Simplify the right side.

Divide $- 49$ by $- 1$ .

$x^{2} \leq 49$

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

$\sqrt{x^{2}} \leq \sqrt{49}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{49}$

Simplify the right side.

Simplify $\sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$| x | \leq \sqrt{7^{2}}$

Pull terms out from under the radical.

$| x | \leq | 7 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | \leq 7$

Write $| x | \leq 7$ as a piecewise.

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

$x \geq 0$

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

$x \leq 7$

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

$x < 0$

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

$- x \leq 7$

Write as a piecewise.

${\begin{cases} x \leq 7 & x \geq 0 \\ - x \leq 7 & x < 0 \end{cases}$

Find the intersection of $x \leq 7$ and $x \geq 0$ .

$0 \leq x \leq 7$

Solve $- x \leq 7$ when $x < 0$ .

Divide each term in $- x \leq 7$ by $- 1$ and simplify.

Divide each term in $- x \leq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{7}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x \geq - 7$

Find the intersection of $x \geq - 7$ and $x < 0$ .

$- 7 \leq x < 0$

Find the union of the solutions.

$- 7 \leq x \leq 7$

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

$[- 7, 7]$

Find the intersection of $y < 0$ and $[- 7, 7]$ .

$- 7 \leq y < 0$

Write as a piecewise.

${\begin{cases} y < \sqrt{(7 + x) (7 - x)} & 0 \leq y \leq 7 \\ - y < \sqrt{(7 + x) (7 - x)} & - 7 \leq y < 0 \end{cases}$

Find the intersection of $y < \sqrt{(7 + x) (7 - x)}$ and $0 \leq y \leq 7$ .

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

Solve $- y < \sqrt{(7 + x) (7 - x)}$ when $- 7 \leq y < 0$ .

Divide each term in $- y < \sqrt{(7 + x) (7 - x)}$ by $- 1$ and simplify.

Divide each term in $- y < \sqrt{(7 + x) (7 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Divide $y$ by $1$ .

$y > \frac{\sqrt{(7 + x) (7 - x)}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{(7 + x) (7 - x)}}{- 1}$ .

$y > - 1 \cdot \sqrt{(7 + x) (7 - x)}$

Rewrite $- 1 \cdot \sqrt{(7 + x) (7 - x)}$ as $- \sqrt{(7 + x) (7 - x)}$ .

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

Find the intersection of $y > - \sqrt{(7 + x) (7 - x)}$ and $- 7 \leq y < 0$ .

No solution

Find the union of the solutions.

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

Solve for $y$ .

Subtract $x^{2}$ from both sides of the inequality.

$y^{2} > 25 - x^{2}$

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

$\sqrt{y^{2}} > \sqrt{25 - x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | > \sqrt{25 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{25 - x^{2}}$ .

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

$| y | > \sqrt{5^{2} - x^{2}}$

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 $b = x$ .

$| y | > \sqrt{(5 + x) (5 - x)}$

Write $| y | > \sqrt{(5 + x) (5 - x)}$ 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.

$y > \sqrt{(5 + x) (5 - x)}$

Find the domain of $y > \sqrt{(5 + x) (5 - x)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y > \sqrt{(5 + x) (5 - x)}$ .

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

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

Solve for $y$ .

Simplify $(5 + x) (5 - x)$ .

Expand $(5 + x) (5 - x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$25 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Multiply $- 1$ by $5$ .

$25 - 5 x + x \cdot 5 + x (- x) \geq 0$

Move $5$ to the left of $x$ .

$25 - 5 x + 5 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$25 - 5 x + 5 x - x \cdot x \geq 0$

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

Move $x$ .

$25 - 5 x + 5 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$25 - 5 x + 5 x - x^{2} \geq 0$

Add $- 5 x$ and $5 x$ .

$25 + 0 - x^{2} \geq 0$

Add $25$ and $0$ .

$25 - x^{2} \geq 0$

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 25}{- 1}$

Simplify the right side.

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Simplify the right side.

Simplify $\sqrt{25}$ .

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

$| x | \leq \sqrt{5^{2}}$

Pull terms out from under the radical.

$| x | \leq | 5 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Write $| x | \leq 5$ as a piecewise.

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

$x \geq 0$

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

$x \leq 5$

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

$x < 0$

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

$- x \leq 5$

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Solve $- x \leq 5$ when $x < 0$ .

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Simplify the right side.

Divide $5$ by $- 1$ .

$x \geq - 5$

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Find the union of the solutions.

$- 5 \leq x \leq 5$

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

$[- 5, 5]$

Find the intersection of $y \geq 0$ and $[- 5, 5]$ .

$0 \leq y \leq 5$

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 $- 1$ .

$- y > \sqrt{(5 + x) (5 - x)}$

Find the domain of $- y > \sqrt{(5 + x) (5 - x)}$ and find the intersection with $y < 0$ .

Find the domain of $- y > \sqrt{(5 + x) (5 - x)}$ .

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

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

Solve for $y$ .

Simplify $(5 + x) (5 - x)$ .

Expand $(5 + x) (5 - x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$25 + 5 (- x) + x \cdot 5 + x (- x) \geq 0$

Multiply $- 1$ by $5$ .

$25 - 5 x + x \cdot 5 + x (- x) \geq 0$

Move $5$ to the left of $x$ .

$25 - 5 x + 5 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$25 - 5 x + 5 x - x \cdot x \geq 0$

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

Move $x$ .

$25 - 5 x + 5 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$25 - 5 x + 5 x - x^{2} \geq 0$

Add $- 5 x$ and $5 x$ .

$25 + 0 - x^{2} \geq 0$

Add $25$ and $0$ .

$25 - x^{2} \geq 0$

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 25}{- 1}$

Simplify the right side.

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Simplify the right side.

Simplify $\sqrt{25}$ .

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

$| x | \leq \sqrt{5^{2}}$

Pull terms out from under the radical.

$| x | \leq | 5 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Write $| x | \leq 5$ as a piecewise.

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

$x \geq 0$

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

$x \leq 5$

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

$x < 0$

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

$- x \leq 5$

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Solve $- x \leq 5$ when $x < 0$ .

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Simplify the right side.

Divide $5$ by $- 1$ .

$x \geq - 5$

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Find the union of the solutions.

$- 5 \leq x \leq 5$

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

$[- 5, 5]$

Find the intersection of $y < 0$ and $[- 5, 5]$ .

$- 5 \leq y < 0$

Write as a piecewise.

${\begin{cases} y > \sqrt{(5 + x) (5 - x)} & 0 \leq y \leq 5 \\ - y > \sqrt{(5 + x) (5 - x)} & - 5 \leq y < 0 \end{cases}$

Find the intersection of $y > \sqrt{(5 + x) (5 - x)}$ and $0 \leq y \leq 5$ .

No solution

Solve $- y > \sqrt{(5 + x) (5 - x)}$ when $- 5 \leq y < 0$ .

Divide each term in $- y > \sqrt{(5 + x) (5 - x)}$ by $- 1$ and simplify.

Divide each term in $- y > \sqrt{(5 + x) (5 - x)}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} < \frac{\sqrt{(5 + x) (5 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} < \frac{\sqrt{(5 + x) (5 - x)}}{- 1}$

Divide $y$ by $1$ .

$y < \frac{\sqrt{(5 + x) (5 - x)}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{(5 + x) (5 - x)}}{- 1}$ .

$y < - 1 \cdot \sqrt{(5 + x) (5 - x)}$

Rewrite $- 1 \cdot \sqrt{(5 + x) (5 - x)}$ as $- \sqrt{(5 + x) (5 - x)}$ .

$y < - \sqrt{(5 + x) (5 - x)}$

Find the intersection of $y < - \sqrt{(5 + x) (5 - x)}$ and $- 5 \leq y < 0$ .

$- 5 \leq y < N o (Min i m u m)$

Find the union of the solutions.

$- 5 \leq y < i N o Min m^{2} u$

Plot each graph on the same coordinate system.

$x^{2} + y^{2} < 49$

$x^{2} + y^{2} > 25$

Graph x^2+y^2<49 x^2+y^2>25

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