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Algebraic formulas

  1. Feb 19, 2006 #1
    I have often noticed something between distance formula:
    [tex] d(P_1,P_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

    and equation of a circle:


    There appears to be a relation between the two. It seems as though both [tex]h,k[/tex] can be replaced with an [tex]x,y[/tex] (in the eqaution of a circle formula) and then placed in a radical to determine the square root of the radius, or the in other words the distance.
    Is there some relation between the two, and why isn't it discussed in algebra courses? I think there is, but my high school teachers never touched on it and neither did my algebra professor.
    Last edited: Feb 19, 2006
  2. jcsd
  3. Feb 19, 2006 #2


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    A circle of radius r centered at (h,k) is the set of points whose distance is r from (h,k).
  4. Feb 20, 2006 #3


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    It's certainly is discussed in every course I've seen! As AKG pointed out, its because a circle is defined as the set of points a fixed distance from the center. Typically, the derivation of the equation for a circle is done by noting that, if a circle has center (a,b) and radius r, then
    [tex]\sqrt{(x-a)^2+ (y-b)^2}= r[/tex]
    and squaring both sides.
    (Edited thanks to VietDao29)
    Last edited: Feb 20, 2006
  5. Feb 20, 2006 #4


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    Nope, in fact, it should read:
    [tex]\sqrt{(x - a) ^ 2 + (y - b) ^ 2}= \sqrt{r ^ 2} = r[/tex] (r > 0)
    You forgot a square root. :)
  6. Feb 20, 2006 #5


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    It might be that the professor in Plastic Photon's course thought it too trivial to mention.
    However, I'd like to give Plastic Photon the credit for actually thinking about and relating together the formulas he has learned. That is an important step in learning maths.

    Keep the good work up, Plastic Photon! :smile:
  7. Feb 20, 2006 #6


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    Okay, I'll stop being curmudgeonly and say, I, too, was impressed that Plastic Photon realized that without having been told!
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