Posted by Connor on Tuesday, August 31, 2010 at 7:31pm.
Complete the square to find the center and the radius of the circle:
3x^2+3y^212x+24y+15=0

Pre Calc  Reiny, Tuesday, August 31, 2010 at 7:55pm
3x^2+3y^212x+24y+15=0
3(x^2  4x + ...) + 3(y^2 + 8y + ...) = 15
3(x^2  4x + 4) + 3(y^2 + 8y + 16) = 15 + 12 + 48
3(x2)^2 + 3(y+4)^2 = 45
(x2)^2 + 3(y+4)^2 = 9
take it from there.
(I suppose I could have divided each term by 3 at the start) 
Pre Calc  Connor, Tuesday, August 31, 2010 at 7:59pm
Thank you so much, i'm so stressed if its not too much trouble check out some of my other questions;)

Pre Calc  Henry, Tuesday, August 31, 2010 at 8:43pm
3x^2 + 3y^2  12x + 24y + 15 = 0.
Divide both sides of Eq by 3 to reduce
the coefficients of x^2 and y^2 to one:
x^2 + y^2  4x + 8y + 5 = 0
Complete the squares:
x^2 4x + (4/2)^2 + y^2 + 8y +(8/2)^2,
The terms that were added to complete
the square should be added to rt side
also:
x^2 4x + 4 + y^2 +8y +16 = 4 +16  5,
Write the perfect squares as binomials
squared:
(x  2)^2 + (x + 4)^2 = 15,
(x  h)^2 + (y  k)^2 = r^2 = STD Form.
C(h , k) = C(2 , 4), r^2 = 15, r =
sqrt(15) = 3.89. 
Thanks Henry  Pre Calc  Reiny, Tuesday, August 31, 2010 at 9:15pm
Of course 45รท3 = 15 and not 9 like I had.
Silly me!