Question

I have NO idea what to do here: 3x^2+3y^2-12x+24y-48=0

Answer 1

what do they ask you to do?

Answer 2

I believe this is going to be a "completing the square" problem. But right off the bat, I see that every term has a "3" as a factor, so I would start by dividing out a 3 from the whole left side.

Answer 3

this is one equation, with two unknowns. what are you solving for?

Answer 4

Find the center and radius of the following circle

Answer 5

find the center and radius of the circle is my guess

Answer 6

divide everything by 3 is the first step, and fortunately all the terms are divisible by 3
what do you get?

Answer 7

Such a pro, he doesn't even need a question.

Answer 8

ikr, just divine inspiration i guess...

Answer 9

@lizzy89 what do you get when you divide each term by 3?

Answer 10

x^2+y^2-4x+8y-16=0?

Answer 11

I'm not good at this at all!

Answer 12

yeah that looks good

Answer 13

\[3x^2+3y^2-12x+24y-48=0\]
\[x^2+y^2-4x+8y-16=0\]

Answer 14

now the goal it to make this look like
\[(x-h)^2+(y-k)^2=r^2\] so we can read off the answer you are looking for
lets group the terms together

Answer 15

\[x^2+y^2-4x+8y-16=0\]
\[x^2-4x+y^2+8y=16\] is a start
the next part is not too hard either

Answer 16

half of \(4\) is \(2\) and \(2^2=4\) so we can complete the square for the \(x\) term by writing
\[(x-2)^2+y^2+8y=16+4=20\] now we do the same thing for the \(y\) term

Answer 17

half of \(8\) is \(4\) and \(4^2=16\) we can write
\[(x-2)^2+(y+4)^2=20+16=36\]

Answer 18

that is the "completing the square" part
now that you have
\[(x-2)^2+(y+4)^2=36\]it looks just like \[(x-h)^2+(y-k)^2=r^2\] the circle with center \((h,k)\) and radius \(r\)
read the answer directly from the equation
let me know what you get

Answer 19

2 and -4?

Answer 20

6?

Answer 21

@satellite73 I think you mean (2,-4)?

Answer 22

yeah i mean \((2,-4)\) right and \(r=\sqrt{36}=6\)

Answer 23

sorry about that mistake

Answer 24

you were right, i was lunching

Answer 25

Thank you @satellite73 !!

Answer 26

yw