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Homework Help: Circle problem

  1. Sep 20, 2004 #1
    My Calculus math teacher often likes to give us math problems that we have never encountered , for home work , before he actually starts teaching it. According to him it gets us 'thinking' analyticaly :confused: . Anyways here's the problem that's gotten me stomped :

    Find the equation of the circle passing through the points A (0,8) , B (0,6) , and C (0,0).

    Here's what I've figured so far , points A and B forms a chord. I found the mid point of A and B which is (3,4). Then using the point of slop formula , I obtain a line for the diamater of the circle which is y= 3/4 X + 2.25 . At this point I can't figure out what to do next to obtain the radius or the center of the circle. :uhh:

    Some help please? :cry:
     
  2. jcsd
  3. Sep 20, 2004 #2
    How can the midpoint of A and B be (3,4)? I think you made a typo for your question. There are no circles that can pass through (0,8), (0,6) and (0,0). :smile:
     
  4. Sep 20, 2004 #3
    Oops! Sometimes I type too fast and make silly errors :yuck: . It's suppose to be (6,0) for point B NOT (0,6). :tongue2:
     
  5. Sep 20, 2004 #4
    ok
    i got
    center of the circle at 3,4
    with a radius of 5
     
  6. Sep 20, 2004 #5
    It will all make 3,4,5 triangles in this fashion :)
     
  7. Sep 20, 2004 #6
    Algerbra way--the way i did the first time... silly me

    Here is the algerbra way of doing it
    General Form for circle equation
    [tex](x-x_0)^2+(y-y_0)^2=r^2 [/tex]

    Here are the three versions of the equations

    [tex]1. -x_0^2+(8-y_0)^2=r^2 [/tex]
    [tex]2. (6-x_o)^2 + -y_0)^2 = r^2 [/tex]
    [tex]3. -x_0^2+y_0^2=r^2 [/tex]

    multiply equation three by negitive one
    then add it to equation two to get
    [tex]12x_0=36[/tex]
    therefore
    [tex]x_0=3[/tex]

    again use the equation three and multiply it by negitive one
    then add it to equation one to get
    [tex]64=16y_0[/tex]
    therefore
    [tex]y_0=4[/tex]

    then
    simply
    9+16=25
    squareroot of 25
    equals 5
     
  8. Sep 20, 2004 #7

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Clearly the three points are the vertices of a right triangle and, therefore, the hypotenuse is a diameter (twice the radius!) of the circle so the midpoint of the hypotenuse is the center of the circle. The desired equation follows immediately.
     
  9. Sep 20, 2004 #8
    Ahhh thanks , but I've already solved it , well in a different way....

    Using the point slope formula I got the equation for the diameter line Y=3/4X + 1 3/4. Then found the mid point between points AC and BC , then used point slop formula to get diameter line Y=4 for AC and X=3 for BC. I then simply prove that 3,4 is the center by setting y=4 and x=3 equal to y=3/4x +1 3/4 :

    y=4,
    4=3/4x + 1 3/4
    -3/4x = 2.25
    X = 3

    x=3
    y=3/4(3) + 1 3/4
    y= 4

    After that simply plug (3,4) and either point A , B , C back into the circle equation to get the radius.

    Btw , how does the algebra way works? I don't understand it , does it involves something with triangles?
     
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