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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
    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

    again use the equation three and multiply it by negitive one
    then add it to equation one to get

    squareroot of 25
    equals 5
  8. Sep 20, 2004 #7


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    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 :

    4=3/4x + 1 3/4
    -3/4x = 2.25
    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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