Find center and radius of circle if: x^2+8x+6y=-16-y^2. Please help me understand this! I don't know how to make it into standard form.

Question

Find center and radius of circle if: x^2+8x+6y=-16-y^2.   Please help me understand this!  I don't know how to make it into standard form.

phi · Answer

do you know how to complete the square for
x^2 + 8x

anonymous · Answer

No

cwrw238 · Answer

x^2 + y^2 + 8x + 6y + 16 = 0
coup;été the square for x^ + 8x and y^2 + 6y
and convert it into the form 
(x - a)^2 + (y - b)^2 = r^2

phi · Answer

this talks about how to complete the square http://www.khanacademy.org/math/algebra/quadtratics/v/completing-perfect-square-trinomials

anonymous · Answer

ok I'm downloading the video, thanks.

phi · Answer

you start with
x^2+8x+6y=-16-y^2

or 
x^2+8x+y^2+ 6y=-16

I'll complete the square for the x terms:   take 1/2 the 8 from 8x  to get 4. square it to get 16
add 16 -16 (that is zero, so it does not change the equation)
x^2 + 8x +16   -16 + y^2 +6y = -16
now rewrite x^2 + 8x +16 as (x+4)^2  to get

(x+4)^2 -16 + y^2 +6y = -16
we can move the -16 on the left side to the right side (add +16 to both sides) to get
(x+4)^2 + y^2 +6y = 0

compete the square for the y's, and you will get the equation of a circle in standard form