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Math problem about circles, graphs

  1. Jul 28, 2008 #1
    this is the question...

    For what values of k is the graph of the equation x^2 + y^2 +2x-4y+26=k^2 - 4k

    a. a circle
    b. a point
    c. an empty set

    i think i should change the equation to this form...(x+1)^2+(y-2)^2=k^2 - 4k - 29

    then what?
     
  2. jcsd
  3. Jul 28, 2008 #2
    You already got your equation into the form of the equation of a circle. What would cause it to have no solutions? What would cause it to have only 1 solution?
    Hint:Can the left side ever be negative?
     
  4. Jul 28, 2008 #3
    i know that if r is negative...its an empty set.but how do i get k?

    if i input k^2 - 4k - 29 = 0 in the calculator.... k = 7.74 & -3.74.... whats that?
     
  5. Jul 28, 2008 #4
    r = 0 so that the eqn would become a pt.
    then k = 7.74 & -3.74

    how then do i know which numbers would make the eqn a circle/ empty soln?
     
  6. Jul 28, 2008 #5
    oh! i think i got it!

    a. -3.74>k or k>7.74
    b. k = 7.74 or k=-3.74
    c. -3.74>k<7.74

    is this right?
     
  7. Jul 28, 2008 #6
    (x+1)^2+(y-2)^2 = k^2 - 4k - 29

    For empty set, make the right hand side quadratic equation negative. For a point make the right hand side turn out as 0 and for a circle make it turn out as anything else.

    FOR A POINT ---> k^2 - 4k - 29 = 0
    k = 4 +- sqrt (16 + 116) / 2 = 4+- rt (132) / 2 = 4 +- 11.489 / 2 =
    7.7445 or -3.7445
     
  8. Jul 28, 2008 #7
    FOR EMPTY SET :
    k^2 - 4k - 29 < 0
    (k - 7.7445)( k + 3.7445 ) < 0
    k E (-3.7445 , 7.7445) ................ i have given the interval for k.
     
  9. Jul 28, 2008 #8
    and of course , the remaining values of k are for A CIRCLE

    ie. k < -3.7445 and k > 7.7445
     
  10. Jul 28, 2008 #9
    hancyu, you are correct :)
     
  11. Jul 28, 2008 #10
    Good work!
     
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