Question

consider the following equation of a circle.

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

step 1 of 3: find the center, $(h,k)$.

answerhow to enter your answer (opens in new window) 2 points

Explanation:

Step1: Rewrite x - terms

Complete the square for the $x$ - terms in the equation $x^{2}+y^{2}+6x - 6y-18 = 0$. The $x$ - terms are $x^{2}+6x$. Using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, for $x^{2}+6x$, we have $a = x$ and $2ab=6x$, so $b = 3$ and $x^{2}+6x=(x + 3)^{2}-9$.

Step2: Rewrite y - terms

Complete the square for the $y$ - terms. The $y$ - terms are $y^{2}-6y$. Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, for $y^{2}-6y$, we have $a = y$ and $2ab = 6y$, so $b = 3$ and $y^{2}-6y=(y - 3)^{2}-9$.

Step3: Rewrite the circle equation

Substitute the completed - square expressions into the original equation:
\[

$$\begin{align*} x^{2}+y^{2}+6x - 6y-18&=0\\ (x + 3)^{2}-9+(y - 3)^{2}-9-18&=0\\ (x + 3)^{2}+(y - 3)^{2}&=36 \end{align*}$$

\]
The standard form of a circle equation is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle. Comparing $(x + 3)^{2}+(y - 3)^{2}=36$ with the standard form, we have $h=-3$ and $k = 3$.

Answer:

$(-3,3)$