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Solving a circle equation for Y

  1. Aug 10, 2008 #1
    Hello guys,
    I am going into my senior year of high school and i am taking AP calc. I have always been really great at math, and i dont know if its cause its the middle of the summer, or what, but this one problem is giving me some serious trouble. Anyway, the book says to graph the equation, (it might only be possible to solve for y and then graph both equations).

    Equation= 4x2+4Y2-4x+24y-63=0

    I have tried a few times but can't get it to being Y=......

    Any help appreciated (I would like if you could, to show the steps rather than just the answer, because i can get it using my calculator, but i want to be able to actually do it)

  2. jcsd
  3. Aug 11, 2008 #2
    To graph a circle, you should get it in the form (x-a)^2 + (y-b)^2 = r^2. Then you know that it has centre (a,b) and radius r. There is not need to solve in the form you mentioned.
    Now to get to the form (x-a)^2 + (y-b)^2 = r^2, you do some completing the square (which you probably have learned when you study quadratic equations). Let us know if you need more hints.
  4. Aug 11, 2008 #3
    Yenchin, thank you for replying, but i am not sure what you are talking about. For a graph, lets say, using a graphing calculator, there is y=... then the equation. So i would need to get it to y= form. I understand the completing the square thing, but i dont get why the textbook says to solve for y if it is not necessary. When i use my calculator to simplify the equation it comes up with.


    then it also has the same equation without the negative in front of it.

    So, even if this would not be the right way to go about graphing it, how would you solve the previous equation in my first post, to get it equal to this?
  5. Aug 11, 2008 #4
    You won't get it in y= form. The closest you can get is what your calculator gave you, one equation with a minus out front and then the exact same one without the minus.

    The reason being that in y = form you can only graph functions (remember the vertical line test?), whereas a circle isn't a function.

    Soo... if you were to complete the square, you'd get the equation for a circle, letting you, a person, graph it easily, even if a calculator can't.

    If you really need it in y= form, though, you have to go with what your calculator said.
  6. Aug 11, 2008 #5
    So is the calculator giving me a false answer?
    Or is that right, just not an actual graph... If it is a correct way of writing it in Y form, culd someone please show me in steps how to get there?
    Each thime so far that i have tried to go from one to the other one of the signs messes up and i'll get every bit right, except instead of +24y, i'll get like -24y, and its really irritating.
  7. Aug 11, 2008 #6


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    Homework Helper

    The calculator isn't wrong. Not all graphs can be represented in the form y=f(x) without omitting something in the equation f(x,y) = constant. In your case it appears that the calculator can give you the equation either as y=-sqrt(f(x)) or y=sqrt(f(x)). I think this is due to something known as the implicit function theorem? Which is why the best way to represent an equation of a circle/ellipse/quadric surface is to represent it in the various general equations for their graphs.
  8. Aug 11, 2008 #7
    There is nothing wrong with it: y= +/- (sqrt{-4*X2+4*X+99}+6)/2 is correct. However, you have to express it as 2 separate functions,


    on your calculator to get it on your graphing calculator if you need an idea of how the graph looks like. But doing this is more cumbersome and unnecessary. And on top of that, I suspect the circle becomes discontinuous near the x-axis if you plot it on your calculator.

    What your book is saying, for presentation, a simpler algebraic method, and without the use of a calculator*, you can express it in the form:

    (x-a)^2 + (y-b)^2 = r^2

    *This becomes obvious if the question asks you to plot a graph where, say, the coefficients of x and y are real constants that are not known, a, b. Then your graphing calculator is only able to provide you an idea of the shape of the graph (e.g. by assuming a=1, b=1, you get an idea of how it looks)

    Last edited: Aug 11, 2008
  9. Aug 11, 2008 #8


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    Science Advisor

    ephedyn, in the "spoiler" can very close to giving you the answer:

    Complete the square in y (ephedyn did it in both x and y which isn't really necessary):
    4(x- 1/2)2+ 4(y- 3)2= 25
    4(y-3)2= 25- 4(x- 1/2)2
    (or, without completing the square in x, 4(y-3)2= 26- 4x2- 4x)
    [tex](y- 3)^2= \frac{25- 4(x-\frac{1}{2}}{4}[/tex]
    [tex]y- 3= \pm \frac{\sqrt{25- 4(x-\frac{1}{2}}}{2}[/tex]
    and finally
    [tex]y= 3\pm \frac{\sqrt{25- 4(x-\frac{1}{2}}}{2}[/tex]
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