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Point of tangency to a circle from a point not on the circle

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the points of tangency to a circle given by x^2+y^2=9 from point (12,9).


    2. Relevant equations
    dy/dx=-x/y
    (what I've been able to come up so far)


    3. The attempt at a solution
    Taking the derivative I got dy/dx=-x/y

    Let the unknown point of tangency be (a,b)

    y-b=(-a/b)(x-a)
    Simplifying that, I got:
    by-ax=a^2+b^2
    a and b fall on the circle; the circle's equation is x^2+y^2=9; therefore, a^2+b^2=9

    by-ax=9
    (12,9) is a point on this ^ line, so

    9b-12a=9
    b=(4/3)a+1

    Substituting back into the original equation x^2+y^2=9,

    a^2+((4/3)a+1)^2=9

    Simplifying that got me 25a^2+27a-72=0.

    This was the point where I knew I was wrong. Where did I go wrong/how do I fix it?
     
  2. jcsd
  3. Mar 29, 2014 #2

    berkeman

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    Can't you just use trig to find the 2 points? My sketch has two triangles with lots of known information about them...
     
  4. Mar 29, 2014 #3

    HallsofIvy

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    So this is to be the tangent line to the circle at (a, b)?

    No, it's not. Multiplying both sides of y- b= (-a/b)(x- a) gives
    b(y- b)= by- b^2= -ax+ a^2. Now add ax and b^2 to both sides to get
    by+ ax= a^2+ b^2

    by+ ax= 9

    12b+ 9x= 9
    b= (-4/3)ax+ 1

    a^2+((4/3)a+1)^2=9[/quote]
    a^2+ ((-4/3)a+ 1)^2= 9

     
  5. Mar 29, 2014 #4
    So I fixed that and even then, the answer was still way off. Can someone explain how to do this step by step with the final solution?
     
  6. Mar 29, 2014 #5

    SammyS

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    What is the that to which you are referring?

    To give "Step - by - step" is to give the solution.


    Show what your steps are after fixing the that, so that we can help you.
     
  7. Mar 30, 2014 #6

    berkeman

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    How about you post a sketch of the circle and two tangent lines? And label the two right triangles...
     
  8. Mar 30, 2014 #7

    Mentallic

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    As berkeman has said, there will be two tangent lines that pass through an external point to a circle. This means you'll have two values of a, hence two values of b which gives you your points of contact on the circle [itex](a_1,b_1)[/itex] and [itex](a_2,b_2)[/itex].

    And how would you expect the algebra to indicate to you that you have two solutions? Well, quadratics can have two solutions :wink:
     
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