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Alternate ways to show a line is a tangent to a curve

  1. Sep 14, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Show that [itex]x+2y=7[/itex] is a tangent to the circle [itex]x^2+y^2-4x-1=0[/itex]



    2. Relevant equations



    3. The attempt at a solution

    One way would be to solve simultaneously by substituting for x or y and getting a perfect square showing that there is only one point of intersection. Is there any other way to do this? I was thinking about the idea that the angle made by a tangent and a radius is 90 degrees and that the product of the gradients of perpendicular lines is -1. Are such thoughts correct ones?
     
  2. jcsd
  3. Sep 14, 2007 #2
    With calculus you can show that the given line and the line tangent to the circle at the single intersection point are one and the same. But all you need to do is prove that there is only one intersection
     
  4. Sep 14, 2007 #3

    Gib Z

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    Yes they are, however all you are showing is that the straight line is perpendicular to a radius, you arent showing if it actually touches the circle or not, and if it does, how many points.
     
  5. Sep 15, 2007 #4

    rock.freak667

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    So there is no other way to show it is a tangent than only showing that there is only one point of intersection?
     
  6. Sep 15, 2007 #5

    Dick

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    You could also show that the distance to the line on a line from the center of the circle that is normal to your line has length equal to the radius of the circle. But that's the same thing as showing there is only one intersection and takes a lot more words to express.
     
    Last edited: Sep 15, 2007
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