Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^{2}+y^{2}+6x + 6y+14 = 0$
(simplify your answer.)
the center of the circle is $(-3,-3)$.
(type an ordered pair.)
the radius of the circle is $r = 2$.
use the graphing tool to graph the circle.

Explanation:

Step1: Group \(x\) and \(y\) terms

$$
LATEXBLOCK0
$$

Step2: Complete the square for \(x\) terms

For \(x^{2}+6x\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 6x\Rightarrow b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\)

Step3: Complete the square for \(y\) terms

For \(y^{2}+6y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=y\), \(2ab=6y\Rightarrow b = 3\). Then \(y^{2}+6y=(y + 3)^{2}-9\)

Step4: Substitute back into the equation

$$
LATEXBLOCK1
$$

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Answer:

The standard form of the equation is \((x + 3)^{2}+(y + 3)^{2}=4\). The center of the circle is \((-3,-3)\) and the radius \(r = 2\)