Question

find the center and the radius of the circle. 3x^2+3y^2-6x+12y=0

Answer 1

first step is to divide everything by 3 to get
\[x^2+y^2-2x+6y=0\]

Answer 2

then you have to "complete the square" via
\[x^2-2x+y^2+6y=0\]
\[(x-1)^2+(y+3)^2=1^2+3^3=10\]

Answer 3

hope it is clear where the -1 and 3 came from
-1 is half of -2 and 3 is half of 6

Answer 4

now you can read the answer because it is in the form
\[(x-h)^2+(y-k)^2=r^2\] a circle with center (h,k) and radius r

Answer 5

if you dividin everything by 3 wouldn the 12 change too 4?

Answer 6

oh poop i screwed up and you are right!

Answer 7

\[x^2-2x+y^2+4y=0\]
\[(x-1)^2+(y+2)^2=1^2+2^2=5\]

Answer 8

good eye!

Answer 9

center (1,-2) radius
\[\sqrt{5}\] sorry for my mistake, glad you caught it!

Answer 10

thanks