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.
$x^2 + y^2 + 2x + 4y + 4 = 0$
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.
click to enlarge graph

Explanation:

Step1: Group x and y terms

Group \(x\) - terms and \(y\) - terms: \((x^{2}+2x)+(y^{2}+4y)+4 = 0\)

Step2: Complete the square for x

For \(x^{2}+2x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\) and \(2ab = 2x\) (so \(b = 1\)). Add and subtract \(1\): \((x^{2}+2x+1-1)\)

Step3: Complete the square for y

For \(y^{2}+4y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\) and \(2ab=4y\) (so \(b = 2\)). Add and subtract \(4\): \((y^{2}+4y + 4-4)\)

Step4: Rewrite the equation

The equation becomes \((x + 1)^{2}-1+(y + 2)^{2}-4+4=0\)
Simplify to \((x + 1)^{2}+(y + 2)^{2}=1\)

Answer:

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