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 (simplify your answer.)

Explanation:

Step1: Group x - terms and y - terms

$$(x^{2}+2x)+(y^{2}+4y)+4 = 0$$

Step2: Complete the square for x - terms

For the expression $x^{2}+2x$, we add $(\frac{2}{2})^{2}=1$ inside the parentheses. To keep the equation balanced, we also subtract 1 from the left - hand side of the equation.
$$(x^{2}+2x + 1-1)+(y^{2}+4y)+4 = 0$$
$$(x + 1)^{2}-1+(y^{2}+4y)+4 = 0$$

Step3: Complete the square for y - terms

For the expression $y^{2}+4y$, we add $(\frac{4}{2})^{2}=4$ inside the parentheses. To keep the equation balanced, we also subtract 4 from the left - hand side of the equation.
$$(x + 1)^{2}-1+(y^{2}+4y + 4-4)+4 = 0$$
$$(x + 1)^{2}+(y + 2)^{2}-1-4 + 4=0$$

Step4: Simplify the equation

$$(x + 1)^{2}+(y + 2)^{2}=1$$

Answer:

$$(x + 1)^{2}+(y + 2)^{2}=1$$