Consider the circle W x^2+y^2=81 . Let A B be a diameter of circle W . A B is extended through A to C . Point T lies on W so that line C T is tangent to W . Point P is the foot of the perpendicular from A to the line C T . Find the maximum value of (BP) ^2

Step-by-step solution

We have (B P)^2=(B Q)^2+(Q P)^2=(O T+B M)^2+(O M+A R)^2 aligned & =9^2(1+ )^2+(9 +9 )^2=9^2(1+ )^2+9^2 4( )^2 & =9^2 [1+ ^2 +2 +4 ^2 ]=9^2 [1+ ^2 +2 +4-4 ^2 ] & =9^2 [5+2 -3 ^2 ] aligned aligned & =-(81)(3) [ ^2 - 2 3 - 5 3 ]=-(81)(3) [ ( - 1 3 )^2- 1 9 - 15 9 ] & =-(81)(3) [ ( - 1 3 )^2- 16 9 ]=(81)(3) [ 16 9 - ( - 1 3 )^2 ] aligned For maximum value of (B P)^2 , we must have = 1 3 Hence .(B P)^2 |_ =(81)(3) ( 16 9 )=(27)(16)=432