Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interesting problem I thought of

  1. Apr 10, 2012 #1
    First, I apologize for any awkward language--I'm still relatively new to mathematics. This problem seems to be related to number theory as well if it pans out, but I thought it would be better suited here.

    Given a point (x,y), can you always construct a circle such that the center of the circle lies in the coordinates (0,0), its radius being equal to the magnitude of the length (x,y) and a point on the circumference of the circle intersects (for lack of better words) the coordinate (x,y). In a set of real numbers ℝ?

    If so, can you extend this to a point (x,y,z) and a sphere centered around the coordinate (0,0,0) with a radius equal to the magnitude of (x,y,z) in a set of real numbers ℝ?
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2
    If you're asking if, given a point (x,y), can we construct a circle centred at the origin that passes through (x,y)? Certainly! Convert it into polar, and the equation describing our circle is

    [tex]r\left(\theta\right)=\sqrt{x^2+y^2}[/tex]

    , which is quite easily converted into rectangular. Something similar goes for spheres, with spherical coordinates.
     
  4. Apr 10, 2012 #3
    Really? Any point? That is a very interesting conclusion!

    How about for complex numbers and so forth?

    Edit: Now since I think about it, it seems to be very intuitive and clear. As long as the circle has the same radius as the magnitude of the coordinate, then it will reach that point.
     
  5. Apr 10, 2012 #4
    Well, depending on your definition of a circle, not necessarily the origin, though I think the generally accepted definition of a circle allows it to be just a point. For complex numbers ... hmmm ... I'm pretty sure similar arguments hold. However, what's your definition of a circle in the resulting hyperspace? I'm pretty sure that by orienting ourselves correctly in this hyperspace, we can just pick a plane that passes through the origin and the point (x,y) and apply the same arguments, but I'm not perfect at visualizing hyperspace. Same goes for the 6-dimensional space resulting from "complex cartesian space."
     
  6. Apr 10, 2012 #5
    Well that is funny, I was going to ask about hyper-spheres next But you beat me to it.

    If a circle can intersect any point, does that mean we can give any point a hypothetical slope based on a circle and still be able to justify it -- a sort of "meta-slope"? I know it seems to be a bit extreme since points are infinitesimally small-- but I'm just letting my mind wander a bit. :biggrin: There might be a very rigorous definition of a point that leaves that impossible.
     
  7. Apr 10, 2012 #6
    Sort of. If you've studied calculus, you'll know that these slopes you're asking about are the solutions to

    [tex]2\cdot x+2\cdot y\cdot y'=0[/tex]

    And so the slope will be

    [tex]-\dfrac xy[/tex]

    Your question's still a little unclear. We can assign each point this slope, we can also assign each point the slope 0, we can assign each point any sort of slope! This is related to slope fields when solving differential equations.
     
  8. Apr 10, 2012 #7
    Okay, take circles out of it and imagine only a point on the coordinate (x,y). Would it be justified to give a point a slope?

    Edit: Nvm, so a point can be given any slope.. Hmm, that is interesting. So it isn't a definite thing.
     
  9. Apr 10, 2012 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    When you say "give a point a slope", what do you mean by "slope". I.e. what definition of "slope" are you using?
     
  10. Apr 11, 2012 #9
    Oh no, don't pay any attention to what I was saying. I was just talking about a "meta-slope," so to speak. I was thinking about how each point on an explicit/implicit function can be given a slope through taking the limit of Δy/Δx. But now I see it doesn't make sense because any point can be given any slope--only dependent on the function f(x).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook