Question

find the center and radius of the circle represented by the equation below.
$x^{2}+y^{2}+18x + 8y+16 = 0$
answer attempt 1 out of 2
center: ( , )
radius:

Explanation:

Step1: Group x and y terms

$$x^{2}+18x + y^{2}+8y+16 = 0$$

Step2: Complete the square for x - terms

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

Step3: Complete the square for y - terms

For \(y^{2}+8y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\), \(2ab=8y\Rightarrow b = 4\). So \(y^{2}+8y=(y + 4)^{2}-16\)

Step4: Substitute back into the equation

\((x + 9)^{2}-81+(y + 4)^{2}-16+16=0\)
\((x + 9)^{2}+(y + 4)^{2}=81\)

Answer:

Center: \((-9,-4)\)
Radius: \(9\)