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Equation of a circle

  1. May 6, 2008 #1
    x^2 + y^2 + 2x -12y + 12 = 0

    (x-a)^2 + (y-b)^2 = r^2 in that form.

    why doesnt this work.


    (x+1) + (y-6)

    this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36

    so (x+1) + (y-6) + 49 = 0?

    that would make r^2 negative which is impossible, what am i screwing up here?
     
    Last edited: May 6, 2008
  2. jcsd
  3. May 6, 2008 #2

    Vid

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    Your constant term is wrong. The center is right, but I got 5 as the radius. Try again and show your work.
     
  4. May 6, 2008 #3
    accidental post
     
    Last edited: May 6, 2008
  5. May 6, 2008 #4
    i have shown my working, i just tried to factorize in a way that would give me the co-efficients of x and y properly,

    x^2 + y^2 + 2x -12y + 12 = 0

    (x-a)^2 + (y-b)^2 = r^2 in that form.

    why doesnt this work.


    (x+1) + (y-6)

    this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36

    i did (x + 1) because it gives you the 2x and the x^2 of the equation, but it also gives +1, i kept this on the left side of the equation so i ended up with +13 there. i then did the same for (y-6) and ended up with 36+13 on the left side of the equation which is 49.
     
  6. May 6, 2008 #5
    when i multiply out the numbers in the brackets, do i have to put them over the right side of this equation, without changing the sign? as it were
     
  7. May 6, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    which is the same as x^2+ 2x+ y^2- 12y= -12

    ?? Did you mean (x+1)^2+ (y- 6)^2?

    Did you notice that you never do say what equation you wound up with?

    Okay, (x+ 1)^2+ (y- 6)^2= x^2+ 2x+ 1+ y^2- 12y+ 36= what?
    Since, in order to "complete the square" with x^2+ 2x, you have to add 1, and, in order to "complete the square" with y^2- 12y, you have to add 36, you have to add add a total of 37.
    You started with x^2+ 2x+ y^2- 12y= -12. Adding 37 to both sides of that gives
    (x+ 1)^2+ (y- 6)^2= 37-12= 25. You have -12+ 1+ 36, not 12+ 36+ 1, because they are on the right side of the equation.

    The other way you have done this was, with your original equation, x^2+ y^2+ 2x- 12y+ 12= 0 is to recognize that you need a total of 36+ 1= 37 on left side. Since you already have 12, you need to add 37- 12= 25 to both sides:
    x^2+ y^2+ 2x- 12y+ 12+ 25= 25 or (x^2+ 2x+ 1)+ (y^2- 12y+ 36)= (x+1)^2+ (y- 6)^2= 25.
     
  8. May 6, 2008 #7
    cant i do this without completing the square.

    im just going to factorize, and whatever numbers i get move to the right side of the equation. lol.

    that way it works even if i dont understand it
     
    Last edited: May 6, 2008
  9. May 6, 2008 #8
    nvm i understand it now
     
  10. May 6, 2008 #9

    rock.freak667

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    Homework Helper

    OR you could recall that the equation of a circle with centre (-f,-g) and radius,r,[itex]r^2=f^2+g^2-c[/itex] is [itex]x^2+y^2=2fx+2gy+c=0[/itex].
    Then put your equation in that form and you'll get the centre and radius.
     
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