college algebra
posted by Zach .
Identify the center and radius of the circles represented by the following questions
(a) 3x²  √3x+3y²y+ 1/12=0
(b) 12x² + 12y²+42y=0

first one:
divide each term by 3
x^2 √3/3 + y^2 + y/3 = 1/36
add the terms to complete the square
x^2 √3/3 + 1/12 + y^2 + y/3 + 1/36= 1/36 + 1/12 + 1/36
(x  √3/6)^2 + (y  1/6)^2 = 1/12
centre is (√3/6,1/6)
radius is 1/√12 = 1/2√3 = √3/6
You do the second one, it is easier. 
Would you start by dividing everything by 6? then after that I am confused

no, by 12, you want to start the equation with 1x^2

Ok I started by diving by 12 to get
x² + y² + 3/2y = 0
then
x² + y² +3/2 + ____ = 0 + ____
then
(3/2)² = 3/4
(3/4)² = 9/16
and got the equation
x² + y² + (3/2)y + 9/16 = 0 + 9/16
here is where I am confused
x² + [y+(3/4)]² = (3/4)² 
no
after you divide by 12 you should have had
x^2 + y^2 + (7/2)y = 0
to figure out what you have to add to both sides, take half of the middle term, then square that
so 1/2 of 7/2 = 7/4
and 7/4 squared is 49/16
x^2 + y^2 + (7/2)y + 49/16 = 0 + 49/16
x^2 + (y + 7/4)^2 = 49/16
centre: (0,7/4)
radius: 7/4 
the measures of two sides of a triangle are given the perimeter of the triangle is 13x214x+12 find the measure of the third side