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The logic problem faced in linear algebra

  1. Sep 15, 2011 #1
    I have some difficulties to understand the logic of the solution to this problem.

    Q:The graph of a(x2 +y2)+bx+cy +d = 0 is a circle if a 6= 0. Show that there
    is a circle through any three points in the plane that are not all on a line.

    solution:Insisting that the graph of a(x2+y2)+bx+cy+d = 0 passes through the three points
    gives a homogeneous linear system of three linear equations in four variables. This has a
    nontrivial solution, and a = 0 would be the three points all lie on the line bx+cy +d = 0,
    contrary to the hypothesis. So a = 0, as desired.

    if there is some problem of the solution ,pls point it out and explain to me in a detail way.
    Thanks a lot and this is the first time i use this forum!
     
  2. jcsd
  3. Sep 15, 2011 #2
    sorry for that some typing mistake.
    the solution shuld be:solution:Insisting that the graph of a(x2+y2)+bx+cy+d = 0 passes through the three points
    gives a homogeneous linear system of three linear equations in four variables. This has a
    nontrivial solution, and a = 0 would be the three points all lie on the line bx+cy +d = 0,
    contrary to the hypothesis. So a is not equal to 0, as desired.
     
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