Question

I need to find the center and the radius. I've already gotten to this---> (x^2+12x+36)+(y^2-6y+9)=4
Original equation: 2x^2+2y^2+24x-12y-8=0

Answer 1

Nope

Answer 2

?

Answer 3

divided both sides by 2, you get x^2 +y^2+12x-6y-4=0 ok then?

Answer 4

ok ive gotten to this part---> (x^2+12x+36)+(y^2-6y+9)=49, I know i have to find the squared form of the left side but how do i do that

Answer 5

Now I make it perfect square:
\[x^2+12x \color{red}{+36}+y^2-6y\color{blue}{+9}\color{red}{-36}\color{blue}{-9}-4=0\]

Answer 6

ok, you are correct then,
so for x part, you have (x+6)^2
for y part, you have (y-3)^2
the right hand side is 49
--> equation : \((x+6)^2 +(y-3)^2= 7^2\)

Answer 7

can you make conclusion from here??

Answer 8

yea but how do i get the (x+6)^2 and the (y-3)^2

Answer 9

why??? (x+6)^2 = x^2+12x+36
and (y-3)^2 = y^2-6y+9
is it not that?? is it not the goal of make them in perfect square form??