Q. Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). Let P(2 cos α, 2 sin α), 0 < α < π/2 and Q(2 cos β, 2 sin β) be two points such that (α − β) = π/2. Then the point of intersection of AQ and BP lies on :

Correct Answer: x² + y² − 4y − 4 = 0

Explanation

The given circle x² + y² = 4 intersects the x-axis at A(2, 0) and B(−2, 0).

Points P and Q lie on the circle and are given in parametric form as:

P(2 cos α, 2 sin α) and Q(2 cos β, 2 sin β), with α − β = π/2.

Using trigonometric identities:

cos α = −sin β and sin α = cos β

Hence,

P(−2 sin β, 2 cos β)

Equation of line AQ and equation of line BP are formed using point–slope form. Solving these two equations simultaneously gives the coordinates of their point of intersection.

On simplifying the resulting coordinates, it is found that the point always satisfies:

x² + y² − 4y − 4 = 0

Thus, the locus of the point of intersection of AQ and BP lies on the given circle.

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