Q.

Let the circle x2+y2=4 intersect x-axis at the points A(a,0), a>0 and B(b,0).  Let P(2cosα, 2sinα) 0<α<π2 and Q(2cosβ, 2sinβ) be two points such that (α-β)=π2.  Then the point of intersection of AQ and BP lies on :   [2026]

1 x2+y2−4x−4=0  
2 x2+y2−4y−4=0  
3 x2+y2−4x−4y-4=0  
4 x2+y2−4x−4y=0  

Ans.

(2)

Let point of intersection R(h,k)

mBR=mBP⇒kh+2=2sinα2cosα+2⇒kh+2=tanα2

mAR=mAQ⇒kh-2=2sinβ2cosβ-2=sinβcosβ-1=-cotβ2

α2-β2=π4

tan(α2-β2)=tanπ4=1

tanα2-tanβ21+tanα2tanβ2=1

kh+2+h-2k1+(kh+2)(2-hk)=1

⇒k2+h2-4k(h+2)4h+2=1

⇒h2+k2-44k=1

⇒x2+y2-4y-4=0