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.

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