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Is it possible to define this explicitly?

  1. Sep 6, 2012 #1
    (x-3)^2 + (y-2)^2 = 1

    With some rearranging I get

    y(y-4) = -x^2+6x-13

    Is it possible to define this explicitly in terms of y? It looks like a mathematical impossibility to define it explicitly in terms of y but I am hoping there might be some analytical technique for this.
     
  2. jcsd
  3. Sep 6, 2012 #2

    pwsnafu

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    What does "define explicitly" even mean? The formula at the start explicitly defines a circle.
    Do you want a function ##y(x)##? Then it should be easy to see the answer is no: for every x there are two values of y.
     
  4. Sep 6, 2012 #3
    Define the function explicitly in terms of y.

    And okay.
     
  5. Sep 6, 2012 #4

    AlephZero

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    I don't see why not being single-valued means it's impossible to define something explicitly.

    ##(y-2)^2 = \text{something}##
    ##(y-2) = \pm \sqrt{\text{something}}##
    ...
     
  6. Sep 6, 2012 #5
    Well in this case it is more like (y^2-4y) = something. so it is bit more complicated. Who knows, maybe one day some genius might develop a clever theory to solve it.
     
  7. Sep 6, 2012 #6

    pwsnafu

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    This is why I asked what "define explicitly" means.

    :confused: How did you get that?
    Edit: And why not just complete the square?
     
  8. Sep 7, 2012 #7

    Mark44

    Staff: Mentor

    You can solve the first equation for y in terms of x. However, the equation defines a relation between x and y, but doesn't represent a function. Some values of x map to two different y values.

    Having said all that, it's pretty simple to solve for y. Leave the terms in y on one side, and move the terms in x to the other side. Do not expand the (y - 2)2 term. Then take the square root of both sides, remembering to use ± for the two solutions.
     
  9. Sep 7, 2012 #8
    Oh, now I see what Alephzero was saying.
     
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