Question

chapter 10 review
score: 45/100 answered: 5/10
question 6
$x^{2}+y^{2}-6x + 16y - 8 = 0$ is the equation of a circle with center $(h,k)$ and radius $r$
for:
$h=$
and
$k=$
and
$r=$
question help: video post to forum

Explanation:

Step1: Complete the square for \(x\) terms

Group \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 3\))

Step2: Complete the square for \(y\) terms

Group \(y\) - terms: \(y^{2}+16y=(y + 8)^{2}-64\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 8\))

Step3: Rewrite the circle equation

The original equation \(x^{2}+y^{2}-6x + 16y-8=0\) becomes \((x - 3)^{2}-9+(y + 8)^{2}-64-8=0\)
Simplify to \((x - 3)^{2}+(y + 8)^{2}=81\)

Answer:

\(h = 3\)
\(k=-8\)
\(r = 9\)