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 Circle (Posted on 2005-07-28)
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.

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 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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