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.
$x^{2}+y^{2}+6x + 4y+9 = 0$
the equation in standard form is $(x + 3)^{2}+(y + 2)^{2}=4$
(simplify your answer.)
the center of the circle is $(-3,-2)$
(type an ordered pair.)
the radius of the circle is $r=$

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: Extract \(r\) from the given standard - form equation

Given \((x + 3)^{2}+(y + 2)^{2}=4\), comparing with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(r^{2}=4\).
Take the square root of both sides: \(r=\sqrt{4}\).

Answer:

\(r = 2\)