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General Form of Circle Equation

  1. May 4, 2004 #1
    I'm attempting to write the general form of a circle where I'm given three points, not necessary equally distant from each other: (0,0), (0,8), (6,0)

    I need to write the equation of the circle in general form. I've managed to get pretty close to what the answer in the book states, but I'm still off.

    Since these points are not equally distant from each other, you can't use the midpoint formula (x1+x2)/2, (y1+y2)/2 to get the center of the circle. Is there another method I should be using to find the center so I can determine the radius, and then find the general form for the circle?

    I'm not looking for the answer, I know what the book says in the answer key, I'm just curious as to what I'm missing, then I can try to take it from there. Thanks!
  2. jcsd
  3. May 4, 2004 #2


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    the circle equation is
    Plug in each of the three (x,y) coordinate pairs, and you get 3 separate equations. The unknowns are a,b,r. Solve it.
  4. May 5, 2004 #3
    You probably don't need this, but you also have the parametric equations of [tex](x-a)^2+(y-b)^2=r^2[/tex] which can be written as:
    [itex]rSin[t] - a; \ rCos[t] - b[/itex]
    as t goes from 0 to [tex]2\pi[/tex]

    I may be wrong, so somebody double check it for me.
  5. May 5, 2004 #4
    Ok, so for the three points you get:

    [tex](0-a)^2 + (0-b)^2 = r^2[/tex]
    [tex](0-a)^2 + (8-b)^2 = r^2[/tex]
    [tex](6-a)^2 + (0-b)^2 = r^2[/tex]

    But without knowing a & b, how do you solve for r?
  6. May 5, 2004 #5


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    You have three equations abd three unknowns. You can, for instance, eliminate r by equation the first two; this gives you one equation with only a and b. You do the same with the first and third equations and get a different equation with a and b. Then you have two new equations on a and b only, which you can solve.

    Once you have a and b, you can substitute on one of the original three eqns to get r.
  7. May 6, 2004 #6
    Hmmm.. I can see how I can do it for this equation since [tex](0-a)^2 = a^2[/tex], so then you just take the square root of both sides to solve for a. But if you have (for example) [tex](1-a)^2 = 1-2a+a^2[/tex], you can't just solve for a can you? You'd be left with [tex]a^2-2a = -1+-[/tex](rest of equation)
  8. May 6, 2004 #7


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    Play with them a little bit... when you eliminate r from first and second, you also eliminate a^2 and b^2, so you only get a constant plus a term in b. Something similar happens combining the first and last eqns.
  9. May 6, 2004 #8


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    [tex](0-a)^2 + (0-b)^2 = r^2[/tex]
    so (1) [tex]a^2+ b^2= r^2[/tex]

    [tex](0-a)^2 + (8-b)^2 = r^2[/tex]
    so (2) [tex]a^2+ b^2-16b+ 64= r^2[/tex]

    [tex](6-a)^2 + (0-b)^2 = r^2[/tex]
    so (3) [tex]a^2- 12a+ 36+ b^2= r^2[/tex]

    Subtracting 1 from 3, -12a+ 36= 0 or a= 3.

    Subtracting 1 from 2, -16b+ 64= 0 or b= 4.

    The a2+ b2= 9+ 16= r2
    so r= 5.

    Doesn't come any easier than that!
  10. May 6, 2004 #9
    Ah.. I see.. I knew I had to set one equation equal to another and then reduce, but I didn't see the relationship between the three points. I'll crack open the book again when I get off work and try a few more of those problems. Thanks for the guidence!
  11. May 10, 2004 #10
    Ok.. I've reworked this, and I'm just about where I was at. Here's what my book shows as the general form of the circle with the following points (same as before):

    Points: (0,0), (0,8), (6, 0)
    Book Answer: [tex]x^2+y^2-6x-8y=0[/tex]

    When I work this, I get this:







    As you can see, I'm slightly off, but I don't understand what I've done wrong. Any ideas where I strayed?

    Thanks again for your help!
  12. May 11, 2004 #11


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    I notice everyone takes the long way round. Here's a better way: Look at the first two equations. Don't multiply them out, just subtract them. You get 8-b=b, or b=4. Now do the same to equations 1 and 3. You get 6-a=a, or a=3. Plug these two result into equation 1 and you find r=5.
  13. May 11, 2004 #12


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    You've just made a silly mistake here, that's all (the radius squared is 25)
    Besides, note the swift manipulation done by krab.
  14. May 11, 2004 #13
    Ah.. that's what happens when I work on this late at night. For some reason I figured it didn't need to be squared since I had already determined the radius, but it's clear now.

    That is quite a bit simplier manipulation Krab. I've got a second problem I can try this out on, since I don't believe I did it correctly.... I think I need a bigger eraser :)

    Thanks for working through this with me. I appreciate it!
  15. May 11, 2004 #14
    Another interesting method to use would be to find the equations of the lines that go through 2 of the points (only 2 lines are needed). So, for example, you find the equation of the line through (0,0) and (0,8) and the equation of the line (0,0) and (6,0). Then, you use midpoint formula and find the equation of the perpendicular bisector of each line. The interesection of the 2 perpendicular bisectors? The center of the circle.
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