Badiuddin askIITians.ismu Expert
Last Activity: 14 Years ago
Dear Tapasranjan
x2+y2 - 2λr x -a2=0
(x-λr)2 +y2 =a2 + λr2
center of circle( λr,0)
so distance of center from the origin. = λr
so λ1 ,λ2 ,λ3 are in GP
now let a general point (acos θ ,asinθ) on circle x2+y2=a2
so length of tangent from this point to the given circle
= [(acos θ)2+(asinθ)2 - 2λr acos θ -a2]1/2
=[ - 2λr acos θ ]1/2
so given λ1 ,λ2 ,λ3 are in GP
so -2λ1 acos θ,-2λ2 acos θ,-2λ3 acos θ will also be in GP
[ - 2λ1 acos θ ]1/2 , [ - 2λ2 acos θ ]1/2 , [ - 2λ3 acos θ ]1/2 will also be in GP
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