Question

put the equation in standard form.
x² + y² - 2x + 6y + 9 = 0
(x - ?)² + (y - )² =

Explanation:

Step1: Complete the square for x - terms

We have $x^{2}-2x$. Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, for $x^{2}-2x$, $a = x$ and $2ab=2x$, so $b = 1$. Then $x^{2}-2x=(x - 1)^{2}-1$.

Step2: Complete the square for y - terms

We have $y^{2}+6y$. Using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, for $y^{2}+6y$, $a = y$ and $2ab = 6y$, so $b = 3$. Then $y^{2}+6y=(y + 3)^{2}-9$.

Step3: Rewrite the original equation

Substitute the completed - square expressions into the original equation $x^{2}+y^{2}-2x + 6y+9 = 0$.
\[

$$\begin{align*} (x - 1)^{2}-1+(y + 3)^{2}-9+9&=0\\ (x - 1)^{2}+(y+3)^{2}&=1 \end{align*}$$

\]

Answer:

$(x - 1)^{2}+(y-(- 3))^{2}=1$; so the blanks are filled with 1, - 3, 1 respectively.