SOLUTION: What is the radius of the circle given by the equation below? x^2 + 8x = -y^2 - 16y - 16

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the radius of the circle given by the equation below? x^2 + 8x = -y^2 - 16y - 16      Log On


   

Question 696352: What is the radius of the circle given by the equation below?
x^2 + 8x = -y^2 - 16y - 16
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What is the radius of the circle given by the equation below?
x^2 + 8x = -y^2 - 16y - 16
**
Standard form of a circle: %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2, (h,k)=(x,y) coordinates of center, r=radius.
To find the radius of given circle, convert given equation to standard form.
x^2 + 8x = -y^2 - 16y - 16
complete the square:
x^2+8x=-y^2-16y-16
x^2+8x+y^2+16y=-16
(x^2+8x+16)+(y^2+16y+64)=-16+16+64
%28x%2B4%29%5E2%2B%28y%2B8%29%5E2=64
r^2=64
r=8