Let the circle x^2+y^2=4 intersect x-axis at the points A (a ,0 ) ,a0 and B (b ,0 ). Let P (2 cosα ,2 sinα ), 0

Question

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 cosα ,2 sinα ), 0< < 2 and Q (2 cosβ ,2 sinβ ) be two points such that ( - )= 2. Then the point of intersection of AQ and BP lies on :

Accepted Answer

Correct answer: A. (a) Let point of intersection R (h ,k ) m_ BR=m_ BP ⇒ kh+2= 2 sinα2 cosα+2 ⇒ kh+2= tan 2 m_ AR=m_ AQ ⇒ kh-2= 2 sinβ2 cosβ-2= sinβ cosβ-1=- cot 2 2- 2= 4 tan( 2- 2)= tan 4=1 tan 2- tan 21+ tan 2 tan 2=1 kh+2+ h-2k1+ ( kh+2 ) ( 2-hk )=1 ⇒ k^2+h^2-4 k (h+2 ) 4h+2=1 h^2+k^2-44k=1 x^2+y^2-4y-4=0

$Let the circle x^{2} + y^{2} = 4 intersect x -axis at the points A (a, 0), a > 0 and B (b, 0) . Let P (2 cosα, 2 sinα), 0 < α < \frac{π}{2} and Q (2 cosβ, 2 sinβ) be two points such that (α - β) = \frac{π}{2} . Then the point of intersection of AQ and BP lies on :$

Solution

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