Mathematics · Coordinate Geometry - Circles
Let the circle x2 + y2 = 4 intersect the x-axis at the points A(a, 0), a > 0 and B(b, 0). Let P(2 cos(α), 2 sin(α)), 0 < α < π2 and Q(2 cos(β), 2 sin(β)) be two points such that (α - β) = π2. Then the point of intersection of AQ and BP lies on:
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- 1x2 + y2 - 4x - 4y - 4 = 0
- 2x2 + y2 - 4x - 4y = 0
- 3x2 + y2 - 4x - 4 = 0
- 4x2 + y2 - 4y - 4 = 0✓
Correct answer
x2 + y2 - 4y - 4 = 0
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