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Basic question: Is it possible for the equation of a circle to have an xy term?

  1. Jun 4, 2009 #1
    Can this be an equation of a circle [tex]x^2+y^2-2hx-2ky+2mxy+r=0[/tex]
     
  2. jcsd
  3. Jun 4, 2009 #2
    No. The presence of xy denotes a term of the form (Ax + By)2, which does not form any part of the general equation of a circle. It may, however, be a conic of some sort that is simply rotated. You can get cross terms by applying a rotation change of coordinates (x, y) -> (x', y') to the general equation of a conic, x' = Cx - Sy, y' = Sx + Cy, where S= sin(t) and C = cos(t) where t is the angle of rotation. In the case of a circle, the cross terms cancel.
    You can find the general equation of a conic section here.
     
  4. Jun 5, 2009 #3

    HallsofIvy

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    The presence of "xy" indicates rotation of the axes. And a circle looks exactly the same no matter how it is rotated!
     
  5. Jun 5, 2009 #4
    Perhaps, if m = 0.
     
  6. Jun 5, 2009 #5
    I guess it can when m = 0, r < k2 + h2
     
  7. Jun 5, 2009 #6

    HallsofIvy

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    The question in the title was "Is it possible for the equation of a circle to have an xy term" to which the correct answer is "No".

    The question in the text was "Can this be the equation of a circle [itex]x^2+ y^3- 2hx- 2kx+ 2mxy= r[/itex]" to which the answer is "Yes". It is possible with m= 0 (so the equation does not have a an xy term) and [itex]h^2+ y^2< r[/itex], NOT [itex]r< h^2+ k^2[/itex]. If m= 0 and [itex]h^2+ k^2= r[/itex] the graph is the single point (h, k). If m= 0 [itex]h^2+ k^2> r[/itex] there are no (x, y) points satifying the equation.
     
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