Write in Standard Form x^2+y^2-x+8y+16=0

$x^{2} + y^{2} - x + 8 y + 16 = 0$

Step 1

Subtract $16$ from both sides of the equation.

$x^{2} + y^{2} - x + 8 y = - 16$

Step 2

Complete the square for $x^{2} - x$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 1$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 1}{2 \cdot 1}$

Simplify the right side.

Cancel the common factor of $- 1$ and $1$ .

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

$d = \frac{- 1 (1)}{2 \cdot 1}$

Cancel the common factor.

$d = \frac{- 1 \cdot 1}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 1}{2}$

Move the negative in front of the fraction.

$d = - \frac{1}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 1)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$e = 0 - \frac{1}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{1}{4}$

Subtract $\frac{1}{4}$ from $0$ .

$e = - \frac{1}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(x - \frac{1}{2})}^{2} - \frac{1}{4}$

Step 3

Substitute ${(x - \frac{1}{2})}^{2} - \frac{1}{4}$ for $x^{2} - x$ in the equation $x^{2} + y^{2} - x + 8 y = - 16$ .

${(x - \frac{1}{2})}^{2} - \frac{1}{4} + y^{2} + 8 y = - 16$

Step 4

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

${(x - \frac{1}{2})}^{2} + y^{2} + 8 y = - 16 + \frac{1}{4}$

Step 5

Complete the square for $y^{2} + 8 y$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 8$

$c = 0$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{8}{2 \cdot 1}$

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 4}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{4}{1}$

Divide $4$ by $1$ .

$d = 4$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{8^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{64}{4}$

Divide $64$ by $4$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 4)}^{2} - 16$ .

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

Step 6

Substitute ${(y + 4)}^{2} - 16$ for $y^{2} + 8 y$ in the equation $x^{2} + y^{2} - x + 8 y = - 16$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} - 16 = - 16 + \frac{1}{4}$

Step 7

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = - 16 + \frac{1}{4} + 16$

Step 8

Simplify $- 16 + \frac{1}{4} + 16$ .

Find the common denominator.

Write $- 16$ as a fraction with denominator $1$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16}{1} + \frac{1}{4} + 16$

Multiply $\frac{- 16}{1}$ by $\frac{4}{4}$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16}{1} \cdot \frac{4}{4} + \frac{1}{4} + 16$

Multiply $\frac{- 16}{1}$ by $\frac{4}{4}$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16 \cdot 4}{4} + \frac{1}{4} + 16$

Write $16$ as a fraction with denominator $1$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16 \cdot 4}{4} + \frac{1}{4} + \frac{16}{1}$

Multiply $\frac{16}{1}$ by $\frac{4}{4}$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16 \cdot 4}{4} + \frac{1}{4} + \frac{16}{1} \cdot \frac{4}{4}$

Multiply $\frac{16}{1}$ by $\frac{4}{4}$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16 \cdot 4}{4} + \frac{1}{4} + \frac{16 \cdot 4}{4}$

Combine the numerators over the common denominator.

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 16 \cdot 4 + 1 + 16 \cdot 4}{4}$

Simplify each term.

Multiply $- 16$ by $4$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 64 + 1 + 16 \cdot 4}{4}$

Multiply $16$ by $4$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 64 + 1 + 64}{4}$

Simplify by adding numbers.

Add $- 64$ and $1$ .

${(x - \frac{1}{2})}^{2} + {(y + 4)}^{2} = \frac{- 63 + 64}{4}$

Add $- 63$ and $64$ .

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

Write in Standard Form x^2+y^2-x+8y+16=0

Related Questions

Support LCMGCF.com by disabling your adblocker.