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I Family of Circles

  1. Jul 22, 2016 #1
    Supposed 2 circles ##C_1 : x^2 + y^2 + {f_1}x + {g_1}y + h_1 = 0## and ##C_2 : x^2 + y^2 + {f_2}x + {g_2}y + h_2 = 0## through two points.

    A family of circles can be constructed as ##x^2 + y^2 + {f_1}x + {g_1}y + h_1 + k(x^2 + y^2 + {f_2}x + {g_2}y + h_2) = 0##.

    By altering the k, an infinitely number of circles through those two points are obtained. So is any circle existed that it cannot be obtained by the family of circles?
     
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  3. Jul 22, 2016 #2

    BvU

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    Well, for what value of ##k## does one get back ## C_2## ?
     
  4. Jul 22, 2016 #3
    Let ##C = \{C_1, C_2, C_3, ...\}## be the set of equations of circles through those two points.
    Can we get all the elements of ##C## from the family of circles?
     
  5. Jul 22, 2016 #4

    BvU

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    You get back ##C_1## when choosing ##k=0##. But ##C_2## ?
     
  6. Jul 22, 2016 #5
    Um... choosing ##\lim_{k \rightarrow \infty} k##?
     
  7. Jul 22, 2016 #6

    mfb

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    You cannot do that, k has to be a real number. Do the variables have to have the same value for C2 and the class of circles?

    If all the variables apart from k are fixed in the family of circles, then there are certainly circles not included in that family - your family has just one degree of freedom while the set of all circles in a plane has three.
     
  8. Jul 22, 2016 #7
    ##f_1, f_2, g_1, g_2, h_1, h_2## are fixed.

    Do ##x, y## count as degree of freedom?Do ##f, g, h## have relationship to make circle passing through those points?
     
  9. Jul 22, 2016 #8

    BvU

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    No, ##x## and ##y## are subject to the equations. ##f_1, f_2, g_1, g_2, h_1, h_2## are six degrees of freedom, but four of those are needed to let the circles go through the two points ##(x_1,y_1)## and also through ##(x_2, y_2)##. The other two can be found from the radii of ##C_1## and ##C_2##.

    In the family of circles there is one degree of freedom (k). Makes sense, because the centers of the circles are constrained to be located on the perpendicular bisector of the line segment from ##(x_1,y_1)## to ##(x_2, y_2)##.

    But ##C2## isn't in the family...
     
  10. Jul 22, 2016 #9
    How about modify the argument that exclude ##C_2##?
     
  11. Jul 22, 2016 #10

    BvU

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    I am not sure what you mean with that question ?
     
  12. Jul 22, 2016 #11

    mfb

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    No.

    Your family of circles is not necessarily a family of circles, by the way. Many pairs of circles lead to empty sets for the second equation.
     
  13. Jul 22, 2016 #12
    How about now ##C = \{ C_1, C_3, C_4, ...\}##?
    Can we get all the elements of ##C## from the family of circles?

    Sorry. I do not get it.
     
  14. Jul 22, 2016 #13

    mfb

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    As example, consider:
    C1: ##x^2+y^2-4x-4y+7=0##
    C2: ##x^2+y^2+4x+4y+7=0##
    For k=1, how does the corresponding figure look like?
    Ck=1: ##2x^2+2y^2+14=0##. Try to find solutions for that.
     
  15. Jul 23, 2016 #14

    BvU

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    Not sure I like this example: C1 and C2 have no points in common....

    I'm embarking on some real work now (yuck, against my nature) by bringing in another example:

    ##C_1 :\ x^2 + (y-1)^2 = 5\quad ## and ##\ C_2 :\ x^2 + (y+1)^2 = 5\quad ## with ##\ (-2,0)\ ## and ##\ (2,0) \ ## in common.

    For ##k = -1## we get no solution, for ##k\ne -1## we get
    $$C_k:\ x^2 + \left (y- {1-k\over 1+k}\right )^2 = 5 + \left ({1-k\over 1+k}\right )^2$$ if I'm not mistaken.
    Circles through the two common points with center ##\ \left (0,{1-k\over 1+k} \right )\ ##.

    That means we found an infinity of circles trough the two common points, but not all of them:
    we don't have ##C_2## in the family, and got one black sheep (a no solution non-circle) in return.
     
  16. Jul 23, 2016 #15

    mfb

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    Ah, the "through two points" mean they intersect. Okay, then it works, k=-1 as exception (where the equation becomes linear - it has solutions, but not a circle).
     
  17. Jul 23, 2016 #16

    BvU

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    You're right - I made a sign error and overlooked that ##y=\pm {1\over 2}## solves ##C_{-1}##. Funny case.
     
  18. Jul 24, 2016 #17

    disregardthat

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    The family of circles going through the two points in ##C_1 \cap C_2## is given by ##k_1(x^2+y^2 + f_1x+g_1y+h_1) + k_2(x^2+y^2 + f_2x+g_2y+h_2) = 0##, where ##k_1## and ##k_2## are not both zero. You retrieve the same circle whenever ##k_1## and ##k_2## are scaled by some constant ##\lambda \not = 0##. In other words, such a circle uniquely determines a point ##(k_1:k_2)## on the projective line ##\mathbb{P}^1## and vice versa. The degenerate case is the point ##(k_1:k_2) = (1:-1)##, in which case you end up with a line. In this sense, the affine line ##\mathbb{A}^1 \cong \mathbb{P}^1 \backslash \{(1:-1)\}## parametrizes the family of circles going through these two points.
     
  19. Jul 24, 2016 #18
    When I construct a equation of circle passing through ##(-2, 0)## and ##(2, 0)## which is not ##C_2##, that equation can be retrieved by ##C_k##. If we plot the center of all circles passing through two points, I think it must be a straight line.

    By this expression, does it obtain all circles and straight line passing through 2 points?
     
  20. Jul 25, 2016 #19

    BvU

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    How come no one puts me right there ? ## (y+1)^2 - (y-1)^2 = 0\quad## gives y = 0. One single straight line through both points in the example. No 'funny case' at all, just a sensible limiting case.
    Yep, the y-axis in the example. Minus the point ##(-1,0)## :smile:
     
  21. Jul 30, 2016 #20

    disregardthat

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    Yes. The degenerate case can also be thought of as a generalized circle.
     
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