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 ( square ). (simplify your answer.)

Explanation:

Step1: Group \(x\) and \(y\) terms

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

Step2: Complete the square for \(x\) - terms

For the \(x\) - terms \(x^{2}+2x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a=x\) and \(2ab = 2x\) (so \(b = 1\)). Add and subtract \(1\):
\((x^{2}+2x+1-1)+(y^{2}+4y)+4=0\)
\((x + 1)^{2}-1+(y^{2}+4y)+4=0\)

Step3: Complete the square for \(y\) - terms

For the \(y\) - terms \(y^{2}+4y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab=4y\) (so \(b = 2\)). Add and subtract \(4\):
\((x + 1)^{2}-1+(y^{2}+4y + 4-4)+4=0\)
\((x + 1)^{2}-1+(y + 2)^{2}-4 + 4=0\)

Step4: Simplify the equation

Simplify the left - hand side of the equation:
\((x + 1)^{2}+(y + 2)^{2}-1=0\)
\((x + 1)^{2}+(y + 2)^{2}=1\)

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

For the equation \((x + 1)^{2}+(y + 2)^{2}=1=(x-(-1))^{2}+(y-(-2))^{2}=1^{2}\)

The center of the circle \((h,k)=(-1,-2)\) and the radius \(r = 1\)

Answer:

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