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.
$x^{2}+y^{2}+6x + 6y+14 = 0$
the equation in standard form is $(x + 3)^{2}+(y + 3)^{2}=4$.
(simplify your answer.)
the center of the circle is $(-3,-3)$.
(type an ordered pair.)
the radius of the circle is $r = \square$.

Explanation:

Step1: Recall the standard form of a circle equation

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.

Step2: Compare with the given standard - form equation

We have the equation \((x + 3)^2+(y + 3)^2=4\). We can rewrite \(4\) as \(2^2\).

Answer:

The radius of the circle is \(r = 2\)