Question

Please help me, I need to find the radius and center of 3x^2+3y^2 -6x +12y=0. I grouped the x's and y's together and then factored, but now I am lost.

Answer 1

i thin u have to factor a 3 then complete the square

Answer 2

\[3x ^{2} + 3y ^{2} -6x +12y=0\]
\[3x(x-2) + 3y(y+4) = 0\]

Answer 3

I do not know what to do now.

Answer 4

The first step is to divide both sides of the equation by 3

divide each term by 3

Answer 5

you do not factor 3x^2 -6x

you complete the square

Answer 6

I got (x-1) + (y+2)=5

Answer 7

I understand it a little bit now. Thank you.

Answer 8

you keep leaving out the exponents.

Let's see what we get if we expand your answer

(x-1)^2 + (y+2)^2 = 5

x^2 -2x+1 + y^2 +4y +4 -5=0
x^2 -2x +y^2 +4y =0

now if we multiply both sides by 3 we get
3x^2 -6x +3y^2 +12y =0
it matches.

So, except for leaving out ^2 your have the correct answer

Answer 9

what would I do if \[2x ^{2} +2y ^{2}-4x=0 \] was my equation?

Answer 10

since there are only 3 terms, I cannot do what I just did correct?

Answer 11

You would divide both sides of the equation by 2
to get
x^2 +y^2 -2x =0

complete the square for the x's
the y is already a perfect square
(remember, the equation of a circle at (0,0) is x^2 + y^2 = r^2
or if you like (x-0)^2 + (y-0)^2 = r^2

your circle's center will have a y=0, and we have to find the x, by completing the square

Answer 12

(x-1)+(y-0) =1?

Answer 13

I will say correct if you write it this way
(x-1)^2+(y-0)^2 =1

Answer 14

lol (x-1)^2 +(y-0)^2=1 thank you

Answer 15

and people don't bother to write (y-0) I put it there so you can "see" that it matches our pattern.

so on a test you would see a choice like this:
(x-1)^2 + y^2=1

Answer 16

that's good to know

Answer 17

can I ask you one more question?

Answer 18

so your circle has a center at (1,0) and a radius of 1.
We can use wolfram to check the answer
http://www.wolframalpha.com/input/?i=2x%5E2%2B2y%5E2 −4x%3D0

click on properties to the right of where it says circle

Answer 19

Please make your new question a separate post.

Answer 20

The wolfram link needs that extra bit -4x%3D0

Answer 21

thank you