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Analytics of an Arc

  1. Jan 30, 2004 #1
    The first general circle formula is,
    [tex]
    (x-a)^2+(y-b)^2=r^2
    [/tex]
    Where M(a,b) and r:radius.
    I understand this well, but when the subject is arcs...
    [tex]
    (x-a)^2=r^2-(y-b)^2
    [/tex]
    [tex]
    x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}
    [/tex]
    My teacher said that equations for x1 and x2 were half circles at right and left. But how?
    And also the same fo y,
    [tex]
    y_\textrm{1,2}=b(+-)\sqrt{r^2-(x-a)^2}
    [/tex]
    were the arcs of top half and bottom of the circle. But why?
    Any help is appreciated.
     
  2. jcsd
  3. Jan 30, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You apparently accept that [tex](x-a)^2+(y-b)^2=r^2[/tex]
    is the equation of a circle.

    To get [tex]x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}[/tex], you solve for x. Of course, with the square root, you have to take + and - to get both roots.

    You know, I hope, that x measures right and left on a graph. The point (4,3) is 4 units to the right of the x-axis. The point (-4,3) is 4 units to the left. When you solve any equation for x, the result is "left" or "right". Taking the positive sign is right, negative, left.

    y measures up and down so solving for y does the same thing except up and down instead of right and left.
     
  4. Jan 31, 2004 #3
    I knew that basics but couldn't put together. Thanks anyway.
     
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