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Tangent lines to a circle

  1. Aug 20, 2008 #1
    1. The problem statement, all variables and given/known data
    You have a circle with the equation x[tex]^{2}[/tex] + (y + 1)[tex]^{2}[/tex] = 1. You can draw to two tangent lines to that circle that intersect the point (0,1) What are the equations of these lines? And you can't use any calculus, derivatives and the like.


    2. Relevant equations
    y=mx+b
    quadratic formula
    x[tex]^{2}[/tex] + (y + 1)[tex]^{2}[/tex] = 1

    3. The attempt at a solution
    Well you can begin by knowing that the lines y-int will be 1 so y=mx+1. Then you can solve the equation for the circle for y which gives you y = -1 [tex]\pm[/tex] [tex]\sqrt{1-x^{2}}[/tex]. Then you can set that equation equal to 0 and get (after factoring) [tex] (1+m ^{2}) * x^{2} + 4mx + 3 = 0[/tex]. Then you can plug this into the quadratic formula to get your x, but there I get stuck. i try to plug that back into y = mx +1 but I don't know what I am looking for.
     
  2. jcsd
  3. Aug 20, 2008 #2

    rock.freak667

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    Homework Helper

    Tangent only touches the circle right? So if you solve, you should get only one solution.

    For [itex]ax^2+bx+c=0[/itex] if [itex]b^2-4ac=0[/itex] how many solutions are there?
     
  4. Aug 20, 2008 #3
    the data is correct, the point (0,1) does not lie on the circle the highest point of the circle is (0,0)
     
  5. Aug 20, 2008 #4

    Defennder

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    Well you have equation of the tangent lines: y=mx+1. Note that by symmetry, you can assume that the x-coordinate of the point where the lines touch the circle are x and -x and their y coordinates are the same So, you have y=mx+1 for both equations:
    (1)y=m1x + 1
    (2)y=m2(-x) + 1

    You also have the equation of the circle. You should see how to continue from here.

    EDIT: This thread shouldn't be in this forum since you can't use calculus to solve the problem.
     
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