Write in Standard Form 9x^2+9y^2-18x-6y+1=0

$9 x^{2} + 9 y^{2} - 18 x - 6 y + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$9 x^{2} + 9 y^{2} - 18 x - 6 y = - 1$

Step 2

Divide both sides of the equation by $9$ .

$x^{2} + y^{2} - 2 x - \frac{2 y}{3} = - \frac{1}{9}$

Step 3

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

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 ${(x - 1)}^{2} - 1$ .

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

Step 4

Substitute ${(x - 1)}^{2} - 1$ for $x^{2} - 2 x$ in the equation $x^{2} + y^{2} - 2 x - \frac{2 y}{3} = - \frac{1}{9}$ .

${(x - 1)}^{2} - 1 + y^{2} - \frac{2 y}{3} = - \frac{1}{9}$

Step 5

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

${(x - 1)}^{2} + y^{2} - \frac{2 y}{3} = - \frac{1}{9} + 1$

Step 6

Complete the square for $y^{2} - \frac{2 y}{3}$ .

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

$a = 1$

$b = - \frac{2}{3}$

$c = 0$

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

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

Simplify the right side.

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

$e = 0 - \frac{{(- \frac{2}{3})}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $- \frac{2}{3}$ .

$e = 0 - \frac{{(- 1)}^{2} {(\frac{2}{3})}^{2}}{4 \cdot 1}$

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

$e = 0 - \frac{1 {(\frac{2}{3})}^{2}}{4 \cdot 1}$

Apply the product rule to $\frac{2}{3}$ .

$e = 0 - \frac{1 \frac{2^{2}}{3^{2}}}{4 \cdot 1}$

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

$e = 0 - \frac{1 \frac{4}{3^{2}}}{4 \cdot 1}$

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

$e = 0 - \frac{1 (\frac{4}{9})}{4 \cdot 1}$

Multiply $\frac{4}{9}$ by $1$ .

$e = 0 - \frac{\frac{4}{9}}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{\frac{4}{9}}{4}$

Multiply the numerator by the reciprocal of the denominator.

$e = 0 - (\frac{4}{9} \cdot \frac{1}{4})$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = 0 - (\frac{4}{9} \cdot \frac{1}{4})$

Rewrite the expression.

$e = 0 - \frac{1}{9}$

Subtract $\frac{1}{9}$ from $0$ .

$e = - \frac{1}{9}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - \frac{1}{3})}^{2} - \frac{1}{9}$ .

${(y - \frac{1}{3})}^{2} - \frac{1}{9}$

Step 7

Substitute ${(y - \frac{1}{3})}^{2} - \frac{1}{9}$ for $y^{2} - \frac{2 y}{3}$ in the equation $x^{2} + y^{2} - 2 x - \frac{2 y}{3} = - \frac{1}{9}$ .

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} - \frac{1}{9} = - \frac{1}{9} + 1$

Step 8

Move $- \frac{1}{9}$ to the right side of the equation by adding $\frac{1}{9}$ to both sides.

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} = - \frac{1}{9} + 1 + \frac{1}{9}$

Step 9

Simplify $- \frac{1}{9} + 1 + \frac{1}{9}$ .

Combine the numerators over the common denominator.

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} = 1 + \frac{- 1 + 1}{9}$

Simplify the expression.

Add $- 1$ and $1$ .

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} = 1 + \frac{0}{9}$

Divide $0$ by $9$ .

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} = 1 + 0$

Add $1$ and $0$ .

${(x - 1)}^{2} + {(y - \frac{1}{3})}^{2} = 1$

Write in Standard Form 9x^2+9y^2-18x-6y+1=0

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