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Posted by 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 - , 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 - , Wednesday, April 2, 2008 at 1:48pm

    Would you start by dividing everything by 6? then after that I am confused

  • college algebra - , Wednesday, April 2, 2008 at 2:26pm

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

  • college algebra - , Wednesday, April 2, 2008 at 2:29pm

    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

  • college algebra - , 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 - , 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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