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Equation of circle

  1. Nov 25, 2004 #1
    Hi

    we are learning the circle equation now, but I don't understand it at all :confused:

    Please can someone explain the equation in a simple way ?

    thanks alot !

    Roger
     
  2. jcsd
  3. Nov 25, 2004 #2

    arildno

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    Are you talking about:
    [tex]x^{2}+y^{2}=R^{2}[/tex]
    or some different creature?
    Be more specific; post the equation you're confused about!
     
  4. Nov 25, 2004 #3
    Dear Arildno,

    Yes thats the equation which I mean.

    Theres also different variations on it as well with other terms.
    please can you help me


    Roger
     
  5. Nov 25, 2004 #4

    Hurkyl

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    Do you know the geometric definition of a circle?
     
  6. Nov 25, 2004 #5
    the definition of a circle is the locus of all points equidistant from a given point.

    x^2 + y^2 = r^2 is the equation of a circle with radius r, and center at (0,0)

    (x-h)^2 + (y-k)^2 = r^2 is the equation of a circle with center (h,k)

    Then we also have general equation of conic.
     
  7. Nov 25, 2004 #6
    No I don't.

    I don't actually understand the equation or definition ....Thats the problem.

    Arildno mentioned the equation but please can someone explain for me .
    Thanks

    Roger
     
  8. Nov 25, 2004 #7
    A circle can be considered the set of all point that have a certain distance (the radius) from a certain point (the center).

    In above case the center is the origin (0,0) and the distance is R. Thus, the cirlce is the set of all points (x,y) which have a distance R from the origin. The distance of a point (x,y) from the origin is sqrt(x² + y²) using Pythagoras, so your condition for a point to be part of the circle is
    sqrt(x² + y²) = R
    or, when squaring that equation:
    x² + y² = R²
     
  9. Nov 25, 2004 #8

    Math Is Hard

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    Have you tried plotting a "unit circle" on a graph? You can use the equation
    x^2 + y^2 = 1

    if you let y = 0, then you have two possible solutions for x: 1 and -1
    this will give you some coordinates: (1,0) and (-1,0)
    if you let x = 0, then you have two possible solutions for y: 1 and -1
    this will give you some coordinates: (0,1) and (0, -1)

    you can play around with plugging in other numbers between zero and 1 for x and y, and if you plot these, you should see a circle forming on your graph.
    I don't know if this will be helpful to you, but I remember going through this exercise when I studied trig and it helped me visualize how this function worked.
     
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