Write in Standard Form -6x+x^2=97+10y-y^2

$- 6 x + x^{2} = 97 + 10 y - y^{2}$

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

Subtract $10 y$ from both sides of the equation.

$- 6 x + x^{2} - 10 y = 97 - y^{2}$

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

$- 6 x + x^{2} - 10 y + y^{2} = 97$

Move $- 10 y$ .

$- 6 x + x^{2} + y^{2} - 10 y = 97$

Move $- 6 x$ .

$x^{2} + y^{2} - 6 x - 10 y = 97$

Complete the square for $x^{2} - 6 x$ .

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

$a = 1$

$b = - 6$

$c = 0$

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

$d = \frac{- 6}{2 \cdot 1}$

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$d = \frac{2 \cdot - 3}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 3}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 3}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 3}{1}$

Divide $- 3$ by $1$ .

$d = - 3$

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

$e = 0 - \frac{{(- 6)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{36}{4}$

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Subtract $9$ from $0$ .

$e = - 9$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 3)}^{2} - 9$ .

${(x - 3)}^{2} - 9$

Substitute ${(x - 3)}^{2} - 9$ for $x^{2} - 6 x$ in the equation $x^{2} + y^{2} - 6 x - 10 y = 97$ .

${(x - 3)}^{2} - 9 + y^{2} - 10 y = 97$

Move $- 9$ to the right side of the equation by adding $9$ to both sides.

${(x - 3)}^{2} + y^{2} - 10 y = 97 + 9$

Complete the square for $y^{2} - 10 y$ .

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

$a = 1$

$b = - 10$

$c = 0$

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

$d = \frac{- 10}{2 \cdot 1}$

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 5}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 5}{1}$

Divide $- 5$ by $1$ .

$d = - 5$

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

$e = 0 - \frac{{(- 10)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{100}{4}$

Divide $100$ by $4$ .

$e = 0 - 1 \cdot 25$

Multiply $- 1$ by $25$ .

$e = 0 - 25$

Subtract $25$ from $0$ .

$e = - 25$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - 5)}^{2} - 25$ .

${(y - 5)}^{2} - 25$

Substitute ${(y - 5)}^{2} - 25$ for $y^{2} - 10 y$ in the equation $x^{2} + y^{2} - 6 x - 10 y = 97$ .

${(x - 3)}^{2} + {(y - 5)}^{2} - 25 = 97 + 9$

Move $- 25$ to the right side of the equation by adding $25$ to both sides.

${(x - 3)}^{2} + {(y - 5)}^{2} = 97 + 9 + 25$

Simplify $97 + 9 + 25$ .

Add $97$ and $9$ .

${(x - 3)}^{2} + {(y - 5)}^{2} = 106 + 25$

Add $106$ and $25$ .

${(x - 3)}^{2} + {(y - 5)}^{2} = 131$

Write in Standard Form -6x+x^2=97+10y-y^2

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