Let the circle x^ 2 +y^ 2 =4 intersect x-axis at the points A ( a , 0), a 0 and B ( b , 0). Let P (2 , 2 ), 0

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Let the circle x^ 2 +y^ 2 =4 intersect x-axis at the points A ( a , 0), a >0 and B ( b , 0). Let P (2 , 2 ), 0< < 2 and Q (2 , 2 ) be two points such that ( - )= 2 . Then the point of intersection of AQ and BP lies on :

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Correct answer: A. x^ 2 +y^ 2 -4 y-4=0

Let point of intersection R(h,k) m_ BR = m_ BP k h+2 = 2 2 + 2 k h+2 = 2 m_ AR = m_ AQ k h-2 = 2 2 - 2 = - 1 = - 2 2 - 2 = 4 ( 2 - 2 ) = 4 = 1 2 - 2 1 + 2 2 = 1 k h+2 + h-2 k 1 + ( k h+2 ) ( 2-h k ) = 1 k^2 + h^2 - 4 k(h+2) 4 h+2 = 1 h^2 + k^2 - 4 4k = 1 x^2 + y^2 - 4y - 4 = 0

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