Posted by **Zach** on Wednesday, April 2, 2008 at 1:06pm.

Identify the center and radius of the circles represented by the following questions

(a) 3x² - √3x+3y²-y+ 1/12=0

(b) 12x² + 12y²+42y=0

- college algebra -
**Reiny**, Wednesday, April 2, 2008 at 1:33pm
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.

- college algebra -
**Zach**, Wednesday, April 2, 2008 at 1:48pm
Would you start by dividing everything by 6? then after that I am confused

- college algebra -
**Reiny**, Wednesday, April 2, 2008 at 3:11pm
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

- algebra 1 -
**Anonymous**, Friday, April 4, 2008 at 10:37am
the measures of two sides of a triangle are given the perimeter of the triangle is 13x2-14x+12 find the measure of the third side

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