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Line normal to a circle

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A variable line ax+by+c=0, where a,b,c are in A.P (arithmetic progression), is normal to a circle ##(x-\alpha)^2+(y-\beta)^2=\gamma##, which is orthogonal to circle ##x^2+y^2-4x-4y-1=0##. The value of ##\alpha+\beta+\gamma## is equal to
    A)3
    B)5
    C)10
    D)7


    2. Relevant equations
    Condition for orthogonality of two circle: ##2g_1g_2+2f_1f_2=c_1+c_2##


    3. The attempt at a solution
    From the above condition for orthogonality: ##4\alpha+4\beta=\alpha^2+\beta^2-\gamma-1##

    Since the line is normal to the circle, it passes through the centre of circle i.e. ##a\alpha+b\beta+c=0##. What should I do next? :confused:

    I have 2b=a+c but where should I use this?

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. May 9, 2013 #2
    Anyone?
     
  4. May 9, 2013 #3

    Mark44

    Staff: Mentor

    I don't see that you have used the given information that a, b, and c are in arithmetic progression.
     
  5. May 9, 2013 #4
    I know and that's my question. Where am I supposed to use this? Replace c with 2b-a?
     
  6. May 9, 2013 #5

    Mark44

    Staff: Mentor

    Assuming the common difference to be d, you have b = a + d, and c = a + 2d. This makes your line equation
    ax + (a + d) y + a + 2d = 0.

    Solve for y to get y = -a/(a + d) * x + (-a - 2d)/(a + d). This form gives you the slope of the line very easily.
     
  7. May 9, 2013 #6
    I am still a bit lost. Why do we even need the slope here? :confused:
     
  8. May 10, 2013 #7

    haruspex

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    I don't believe there's enough information here. You have 3 equations and 6 unknowns. Now, it could just happen that the one variable of interest is determinable, but I don't see it. Substitute for γ using γ = Z - α-β, substitute for c using c = 2b-a. That gets you to two equations and 5 unknowns. You can use one equation to eliminate one more unknown, and hope that some huge magic cancellation occurs to eliminate 3 more (all except Z), but it doesn't.
     
  9. May 10, 2013 #8
    This is from my test paper and I don't think that it requires too much time solving questions. But there is no guarantee that the question is correct.
    I have the solution booklet too, I don't see from where did they get the values of ##\alpha## and ##\beta##.

    Solution:
    (A part of the solution is on the different page so I am writing down the equations from the first page and attaching the rest of the steps.)
    ##ax+by+c=0##
    ##ax-2b+c=0##
    ##(1,-2)## is the centre of cricle
    attachment.php?attachmentid=58615&stc=1&d=1368169925.jpg
     

    Attached Files:

  10. May 10, 2013 #9

    haruspex

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    Based on that, I think I've figured it out. The question ought to read:
    I suppose that's what they meant by 'variable' line, but it should have been made much clearer.
    All such lines pass through the point (1, -2),
     
  11. May 10, 2013 #10
    That means I can assume any line and continue with that?

    How did the solution even come up with the values of ##\alpha## and ##\beta##?
     
  12. May 10, 2013 #11

    haruspex

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    No - ALL lines of the form ax+by+c=0, where a,b,c are in arithmetic progression. That fixes where the centre of the circle must be; it must be a point that all such lines go through.
     
  13. May 11, 2013 #12
    Thanks haruspex! :smile:

    I assumed two equations, i.e ##x+2y+3=0## and ##y+2=0##, this gave me ##\alpha## and ##\beta##. I substituted them in the equation for orthogonality and found gamma.

    Thank you! :)
     
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