Find the Vertex Form 8x+x^2-2y=64-y^2

$8 x + x^{2} - 2 y = 64 - y^{2}$

Move all terms containing variables to the left side of the equation.

Add $y^{2}$ to both sides of the equation.

$8 x + x^{2} - 2 y + y^{2} = 64$

Move $- 2 y$ .

$8 x + x^{2} + y^{2} - 2 y = 64$

Move $8 x$ .

$x^{2} + y^{2} + 8 x - 2 y = 64$

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

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 ${(x + 4)}^{2} - 16$ .

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

Substitute ${(x + 4)}^{2} - 16$ for $x^{2} + 8 x$ in the equation $x^{2} + y^{2} + 8 x - 2 y = 64$ .

${(x + 4)}^{2} - 16 + y^{2} - 2 y = 64$

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

${(x + 4)}^{2} + y^{2} - 2 y = 64 + 16$

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

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

$a = 1$

$b = - 2$

$c = 0$

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 $d = \frac{b}{2 a}$ .

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

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

${(y - 1)}^{2} - 1$

Substitute ${(y - 1)}^{2} - 1$ for $y^{2} - 2 y$ in the equation $x^{2} + y^{2} + 8 x - 2 y = 64$ .

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

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

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

Simplify $64 + 16 + 1$ .

Add $64$ and $16$ .

${(x + 4)}^{2} + {(y - 1)}^{2} = 80 + 1$

Add $80$ and $1$ .

${(x + 4)}^{2} + {(y - 1)}^{2} = 81$

Find the Vertex Form 8x+x^2-2y=64-y^2

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