Find the x and y Intercepts (x+3)^2+(y+4)^2=25

${(x + 3)}^{2} + {(y + 4)}^{2} = 25$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

${(x + 3)}^{2} + {((0) + 4)}^{2} = 25$

Solve the equation.

Subtract ${((0) + 4)}^{2}$ from both sides of the equation.

${(x + 3)}^{2} = 25 - {((0) + 4)}^{2}$

Add $0$ and $4$ .

${(x + 3)}^{2} = 25 - 4^{2}$

Raise $4$ to the power of $2$ .

${(x + 3)}^{2} = 25 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

${(x + 3)}^{2} = 25 - 16$

Subtract $16$ from $25$ .

${(x + 3)}^{2} = 9$

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

$x + 3 = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$x + 3 = \pm \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + 3 = \pm 3$

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.

$x + 3 = 3$

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$x = 3 - 3$

Subtract $3$ from $3$ .

$x = 0$

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

$x + 3 = - 3$

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$x = - 3 - 3$

Subtract $3$ from $- 3$ .

$x = - 6$

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

$x = 0, - 6$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (- 6, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

${((0) + 3)}^{2} + {(y + 4)}^{2} = 25$

Solve the equation.

Subtract ${((0) + 3)}^{2}$ from both sides of the equation.

${(y + 4)}^{2} = 25 - {((0) + 3)}^{2}$

Add $0$ and $3$ .

${(y + 4)}^{2} = 25 - 3^{2}$

Raise $3$ to the power of $2$ .

${(y + 4)}^{2} = 25 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

${(y + 4)}^{2} = 25 - 9$

Subtract $9$ from $25$ .

${(y + 4)}^{2} = 16$

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

$y + 4 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$y + 4 = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y + 4 = \pm 4$

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.

$y + 4 = 4$

Move all terms not containing $y$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$y = 4 - 4$

Subtract $4$ from $4$ .

$y = 0$

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

$y + 4 = - 4$

Move all terms not containing $y$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$y = - 4 - 4$

Subtract $4$ from $- 4$ .

$y = - 8$

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

$y = 0, - 8$

y-intercept(s) in point form.

y-intercept(s): $(0, 0), (0, - 8)$

Step 3

List the intersections.

x-intercept(s): $(0, 0), (- 6, 0)$

y-intercept(s): $(0, 0), (0, - 8)$

Step 4

Find the x and y Intercepts (x+3)^2+(y+4)^2=25

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