Question

the equation of a circle is given below. identify the radius and the center. then graph the circle.
$x^{2}+y^{2}+4x - 2y - 11 = 0$
radius:
center: ()

Explanation:

Step1: Rearrange the equation

Group \(x\) - terms and \(y\) - terms: \((x^{2}+4x)+(y^{2}-2y)=11\)

Step2: Complete the square for \(x\)

For \(x^{2}+4x\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = x\), \(2ab=4x\Rightarrow b = 2\). Then \(x^{2}+4x=(x + 2)^{2}-4\)

Step3: Complete the square for \(y\)

For \(y^{2}-2y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab = 2y\Rightarrow b=1\). Then \(y^{2}-2y=(y - 1)^{2}-1\)

Step4: Substitute back into the equation

\((x + 2)^{2}-4+(y - 1)^{2}-1=11\), which simplifies to \((x + 2)^{2}+(y - 1)^{2}=16\)

Step5: Identify the center and radius

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x + 2)^{2}+(y - 1)^{2}=16=(x-(-2))^{2}+(y - 1)^{2}=4^{2}\), we get center \((-2,1)\) and radius \(r = 4\)

Answer:

Radius: \(4\)
Center: \((-2,1)\)