Question

question
find the center and radius of the circle represented by the equation below.
$x^{2}+y^{2}+6x + 18y+9 = 0$
answer attempt 1 out of 2
center: ( , )
radius:

Explanation:

Step1: Rearrange and group terms

Group \(x\) - terms and \(y\) - terms: \((x^{2}+6x)+(y^{2}+18y)+9 = 0\)

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

For \(x^{2}+6x\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\) and \(2b = 6\Rightarrow b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\)

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

For \(y^{2}+18y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=y\) and \(2b = 18\Rightarrow b = 9\). Then \(y^{2}+18y=(y + 9)^{2}-81\)

Step4: Substitute back into the equation

\((x + 3)^{2}-9+(y + 9)^{2}-81+9 = 0\)
Simplify to \((x + 3)^{2}+(y + 9)^{2}=81\)
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.

Answer:

Center: \((-3,-9)\)
Radius: \(9\)