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Finding Points of Tangency for the Unit Circle

  1. Dec 21, 2012 #1
    Hi I'm trying to study over break, this isn't an exact quote but its the part of the problem I need help with. Thanks.

    1. The problem statement, all variables and given/known data
    Draw the unit circle and plot the point P=(3,2). Observe there are TWO lines tangent to the circle passing through the point P. Lines L1 and L2 are tangent to the circle at what points?

    2. Relevant equations

    3. The attempt at a solution
    I tried plugging in the point into a point-slope formula, it was kind of a dead end for me.
     
    Last edited by a moderator: Dec 21, 2012
  2. jcsd
  3. Dec 21, 2012 #2

    phinds

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    draw a figure and show what you know
     
  4. Dec 21, 2012 #3

    haruspex

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    Pick a point on the unit circle, (cos α, sin α) say. Write down the equation of the tangent there.
     
  5. Dec 21, 2012 #4

    phinds

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    How about if we get him to draw a diagram first and then help him from there?
     
  6. Dec 21, 2012 #5
    Equation for unit circle is [itex]y^2 + x^2 = 1[/itex]. Equation for line passing through point P=(3,2) is [itex]y - 2 = m(x-3)[/itex].

    These facts might be helpful. They might also fit under that bold region labeled "2. Relevant equations".

    So, the question is this: if you want to find 2 values for x and y such that both equations are satisfied, what can you do with the two equations? :wink:
     
  7. Dec 21, 2012 #6
    Hi everyone, thank you for the input. Let me clarify my situation, I uploaded a diagram; the approach I had been working on was the one Mandelbroth had mentioned. I had [itex]y = mx-m3+2[/itex] and [itex]y^2 + x^2 = 1[/itex], when I plugged the first equation into the second I wasn't sure what to do especially because there were now 3 unknown variables, y, x, and m. That's where I'm at now.
     

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  8. Dec 21, 2012 #7

    haruspex

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    The problem with htat approach is that nowhere have you specified that the line is supposed to be a tangent. Do you know how to write down the generic equation for a tangent to circle at a given point, (cos θ, sin θ) say?
     
  9. Dec 22, 2012 #8

    Mentallic

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    Yes, so if you plug [itex]y=mx-3m+2[/itex] into [itex]x^2+y^2=1[/itex] then you'll have an equation in terms of x and m. It will be a quadratic in x, so if you solve the quadratic for x, you'll have a solution that is in terms of m.

    Now, what you need to do is to understand what the solution to the quadratic is telling you. What does the discriminant in the quadratic formula say?

    There is no problem with that approach. He'll be picking the values of m that are tangents to the circle soon enough.
     
  10. Dec 22, 2012 #9

    symbolipoint

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    Distance from O to point (3,2) uses distance formula.
    Distance from any point on circle to O is 1 for a unit circle.
    Two points on the circle are tangency points, upper may be called B, (x2, y2).
    Angle of triangle at B is right-angle, so POB is right triangle.
    Distance from B to point (3,2) uses Pythagorean Theorem, because you know the other two triangle side lengths.

    Use DISTANCE FORMULA again for B to point (3,2).

    Two equations and unknowns:
    x2+y2=1
    BP=sqrt((3-x)2+(2-y)2)
     
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