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.
$x^2 + y^2 + 6x + 4y + 12 = 0$

the equation in standard form is \boxed{}.
(simplify your answer.)

the center of the circle is \boxed{}.
(type an ordered pair.)

the radius of the circle is $r = \boxed{}$.

use the graphing tool to graph the circle.
\

$$\begin{center}\\includegraphicsscale=0.5{graph.png}\\end{center}$$

click to enlarge graph

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \(x^{2}+6x + y^{2}+4y=-12\)

Step2: Complete the square for x

For \(x^{2}+6x\), take half of 6 (\(= 3\)), square it (\(= 9\)), add to both sides: \(x^{2}+6x + 9+y^{2}+4y=-12 + 9\)

Step3: Complete the square for y

For \(y^{2}+4y\), take half of 4 (\(= 2\)), square it (\(= 4\)), add to both sides: \(x^{2}+6x + 9+y^{2}+4y + 4=-12 + 9+4\)

Step4: Write as squares

Rewrite as \((x + 3)^{2}+(y + 2)^{2}=1\)

Step5: Identify center and radius

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), so center \((h,k)=(-3,-2)\) and radius \(r = 1\) (since \(r^{2}=1\), \(r=\sqrt{1}=1\))

Answer:

The equation in standard form is \((x + 3)^{2}+(y + 2)^{2}=1\).
The center of the circle is \((-3,-2)\).
The radius of the circle is \(r = 1\).