Question

find the standard equation, center, and radius of
6x^2+6y^2+24x+12y=-6
Show work.

Answer 1

you know how to complete the square?

Answer 2

kinda, i dont understand how to do it, so you could say no

Answer 3

first group the terms:

6x^2+24x+6y^2+12y=-6

Answer 4

now complete the square for each variable

Answer 5

how?

Answer 6

\[6(x^2+4x)+6(y^2+2y)=-6\] divde both sides by 6
\[x^2+4x+y^2+2y=-1\]
\[(x^2+4x+(\frac{4}{2})^2)+(y^2+2y+(\frac{2}{2})^2)=-1+(\frac{4}{2})^2+(\frac{2}{2})^2\]

Answer 7

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

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

Answer 8

you can complete the rest i hope or someone else will :)