Question

the equation of a circle is given below. identify the radius and the center. then graph the circle. $x^{2}+y^{2}+8x + 4y+4 = 0$

Explanation:

Step1: Rewrite the equation by grouping x and y terms

We start with the given equation \(x^{2}+y^{2}+8x + 4y+4 = 0\). Group the \(x\)-terms and \(y\)-terms together: \((x^{2}+8x)+(y^{2}+4y)= - 4\)

Step2: Complete the square for the x - terms

For the \(x\)-terms \(x^{2}+8x\), we use the formula \((a + b)^2=a^{2}+2ab + b^{2}\). Here, \(a=x\) and \(2ab = 8x\), so \(b = 4\). Then \(x^{2}+8x=(x + 4)^{2}-16\)

Step3: Complete the square for the y - terms

For the \(y\)-terms \(y^{2}+4y\), using the same formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab=4y\), so \(b = 2\). Then \(y^{2}+4y=(y + 2)^{2}-4\)

Step4: Substitute back into the equation

Substitute the completed square forms into the equation \((x^{2}+8x)+(y^{2}+4y)=-4\):
\[

$$\begin{align*} (x + 4)^{2}-16+(y + 2)^{2}-4&=-4\\ (x + 4)^{2}+(y + 2)^{2}-20&=-4\\ (x + 4)^{2}+(y + 2)^{2}&=16 \end{align*}$$

\]

The standard form of the equation of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Comparing \((x + 4)^{2}+(y + 2)^{2}=16\) with the standard form, we have \(h=-4\), \(k = - 2\) and \(r^{2}=16\), so \(r = 4\)

Answer:

Radius: \(4\)
Center: \((-4,-2)\)