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How the?

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    This is a simple math equation and i am a bit confused on how

    x^2+y^2=25 makes a circle and how y= sqrt 25-x^2 makes a semi circle
     
  2. jcsd
  3. Feb 3, 2009 #2
    If you would solve [tex]x^{2} + y^{2} = 25[/tex] for y, you would not only get [tex]y=\sqrt{25-x^{2}}[/tex], but also something else. What?
     
  4. Feb 3, 2009 #3
    y=5-x
     
  5. Feb 3, 2009 #4
    No, not quite. (Keep in mind that [tex]\sqrt{a^{2}-b^{2}}\neq a-b[/tex].)

    Think of something as simple as e.g. [tex]x^{2}=9[/tex]. There are two different values of x that satisfy this. What values would that be?
     
  6. Feb 3, 2009 #5
    +-3?
     
  7. Feb 3, 2009 #6

    jgens

    User Avatar
    Gold Member

    Well, since the circle is defined as the locus of all points in a plane which are all a fixed and equal distance from the center, it is fairly intuitively obvious that the implicit equation maps to a circle. A more general case of the formula you presented is x^2 + y^2 = r^2 where r is the radius of the circle. Perhaps a general way to derive and illustrate how that equation makes a circle is to examine the circular definition of the trigonometric functions. We know that sin(theta) = y/r and cos(theta) = x/r where r is the radius and (x,y) is the coordinate of the radius' intersection with the circle. I'm also presuming you're familiar with the identity sin^2(theta) + cos^2(theta) = 1. Placing our unit circle definitions in place of the identity yields x^2 + y^2 = r^2 which may intuitively illustrate how it makes a circle.

    Probably a better way of thinking about it is this: given any ordered pair (x,y) satisfying x^2 + y^2 = r^2, (x,y) must always be some fixed distance r from the center; hence, the equation produces the locus of all points on a plane an equal distance from the center. In other words, x^2 + y^2 = r^2 maps to a circle.
     
  8. Feb 3, 2009 #7
    intence, but okay, what about the second equation which i have mentioned
     
  9. Feb 3, 2009 #8
    y = 5-x is a line, how are you getting semicircle
     
  10. Feb 3, 2009 #9

    jgens

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    Gold Member

    y = 5 - x will definately not make a semi-circle. Solve x^2 + y^2 = r^2 for y and it should be fairly clear why it produces a semi-circle if x^2 + y^2 = r^2 creates a circle.
     
  11. Feb 3, 2009 #10
    Exactly. So if you now have [tex]y^{2}=25-x^{2}[/tex], what two values of y do you get?
     
  12. Feb 4, 2009 #11
    y= SqaUREROOT 25-x^2, (this should be a semicircle right?
     
  13. Feb 4, 2009 #12

    Mark44

    Staff: Mentor

    Yes, this is the upper half of the circle. The lower half is y = [itex]-\sqrt{25 - x^2}[/itex]
     
  14. Feb 4, 2009 #13
    What people were trying to tell you was that if [tex] a = b^2 [/tex] then [tex] b = \pm \sqrt{a} [/tex] NOT just [tex] \sqrt{a} [/tex]
     
  15. Feb 4, 2009 #14
    Alright i understand thanks
     
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