Question

Find the center and radius for the circle:
2x ^{2}+12x+2y ^{2}+12y+4=0

Answer 1

\[2x ^{2}+12x+2y ^{2}+12y+4=0\]

Answer 2

I am having trouble remembering the equations for this conic. How would you find the h and k to to solve for the radius, and how what equation do you use to find the center?

Answer 3

The equation of a circle is
\[(x-a)^2+(x-b)^2=r^2\]

Answer 4

The idea is to get in \((x-a)^2\) \((x-b)^2\) forms, we can use completing the square. Try start by factoring the 2 out of x and y

Answer 5

\[2(x^2+6x)+2(y^2+6y)+4=0\]
Then complete the square for
(x^2+6x) and (y^2+6y)

Answer 6

Thanks ^.^

Answer 7

yw

Answer 8

\[(x^2+6x)+(y^2+6y)=36\] is this correct?

Answer 9

Nope, do it carefully you'll get
\[2[(x+3)^2-9]+2[(y+3)^2-9]+4=0\]
Then
\[2 (x+3)^2+2 (y+3)^2-32=0\]