Question

find the center and radius of the circle represented by the equation below.

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

answer attempt 1 out of 2

center: ( , )

radius:

Explanation:

Step1: Rearrange the equation

Group \(x\) - terms and \(y\) - terms: \((x^{2}-8x)+(y^{2}+16y)= - 16\)

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

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

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

For \(y^{2}+16y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\), \(2ab = 16y\Rightarrow b = 8\). Then \(y^{2}+16y=(y + 8)^{2}-64\)

Step4: Substitute back into the equation

\((x - 4)^{2}-16+(y + 8)^{2}-64=-16\)
\((x - 4)^{2}+(y + 8)^{2}-80=-16\)
\((x - 4)^{2}+(y + 8)^{2}=64\)

Answer:

Center: \((4,-8)\)
Radius: \(8\)