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Circle (Posted on 2005-07-28) Difficulty: 5 of 5
C is a circle with center O. AB is a chord not passing through O. M is the midpoint of AB. C' is the circle with diameter OM. T is a point on C'. The tangent to C' at T meets C at P. Show that PA² + PB² = 4 PT².

No Solution Yet Submitted by nilshady    
Rating: 3.6667 (3 votes)

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Solution Vector Solution | Comment 2 of 5 |
 
Let R and r be the radii of circles C and C' respectively.
Place a coordinate system on the figure such that
  Circle C:    x^2 + y^2 = R^2
  Circle C':   (x-r)^2 + y^2 = r^2
Using vector notation <>,
  <OA> = 2*r<i> + s<j>
  <OB> = 2*r<i> - s<j>
  <OO'> = r<i>
  <O'T> = r*cos(theta)<i> + r*sin(theta)<j>
  <OT> = r*[1 + cos(theta)]<i> + r*sin(theta)<j>
  <OP> = x<i> + y<j>
  , where O' is the center of circle C',
    theta places T any where on circle C', and
    4*r*2 + s^2 = x^2 + y^2 = R^2
  <PA> = (2*r - x)<i> + (s - y)<j>
  <PB> = (2*r - x)<i> - (s + y)<j>
  <PT> = (r*[1 + cos(theta)] - x)<i> + [r*sin(theta) - y)]<j>
Since <PT> and <O'T> are perpendicular, <PT>.<O'T> = 0
  0 = r*cos(theta)*(r*[1 + cos(theta)] - x) +
      r*sin(theta)*[r*sin(theta) - y)]
                       or
  x*cos(theta) + y*sin(theta) = r*[1 + cos(theta)]
Therefore,
  |PA|^2 = <PA>.<PA> = (2*r - x)^2 + (s - y)^2
  |PB|^2 = <PB>.<PB> = (2*r - x)^2 + (s + y)^2
  |PA|^2 + |PB|^2 = 4*(R^2 - 2*r*x)
                  = 4*{(r*[1 + cos(theta)] - x)^2 +
                       [r*sin(theta) - y)]^2}
                  = 4*<PT>.<PT>
                  = 4*|PT|^2

 


  Posted by Bractals on 2005-07-28 21:11:16
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