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Homework Help: Need help on circle/triangle question

  1. Oct 30, 2008 #1
    http://www.maths.ox.ac.uk/filemanager/active?fid=2929 [Broken]
    Question 4 on page 12.

    Could someone give me some hints on how to do this? I have no idea how. All I know is that we could draw a line from the center of the circle to Q, which I think would be perpendicular to that tangent. Also if we draw a horizontal line from the center of the circle to the tangent, the angle between them will be theta.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 30, 2008 #2

    HallsofIvy

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    So you have a circle with center at (1,1), radius 1, and a line tangent to it at Q making angle [itex]\theta[/itex] with the x-axis. The angle [itex]\theta[/itex], as shown, is actually with the "backward pointing" x-axis. The angle with positive x-axis, the one used in defining the slope of a line is [itex]\pi- \theta[/itex]. Since [/itex]tan(\pi- \theta)= - tan(\theta)[/itex], the slope of the tangent line is [itex]-tan(\theta)[/itex] (oops- that was one of the things you were supposed to prove!).
    Of course, the radius of the circle, from (1, 1) to Q, is perpendicular to the tangent line. Let Q= (x,y). Then the line from (1,1) to Q has slope (x-1)/(y-1) which must be the negative reciprocal of [itex]-tan(\theta)[/itex] which is [itex]cot(\theta)[/itex]. The equations
    [tex]\frac{y-1}{x-1}= \frac{cos(\theta)}{sin(\theta)}[/tex]
    and
    [tex]x^2+ y^2= 1[/tex]
    should give you what you need.
     
  4. Oct 31, 2008 #3
    Thanks. I managed to do that part. Are the co-ordinates of P

    P [itex] (1 + \csc \theta + \cot \theta, 0)[/itex]?

    I can't do part (ii). For the explanation I'm unsure. All I know is that angle PRO (in the B area) is [itex]\frac{\pi}{2} - \theta[/itex].

    I got [itex]A(\frac{\pi}{4}) = 1 + \sqrt2 -\frac{3\pi}{8}[/itex]. Is that correct?
     
    Last edited: Oct 31, 2008
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