Let the circle x2 + y2 = 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 :
- x2 + y2 - 4x - 4 = 0
- x2 + y2 - 4x - 4y = 0
- x2 + y2 - 4x - 4y - 4 = 0
- x2 + y2 - 4y - 4 = 0
Answer
(4) x2 + y2 - 4y - 4 = 0
Solution
A(2, 0) and B(−2, 0), so AB is a diameter of the circle (centre O).
- = 2, POQ = 90 , so PAQ = 12 POQ = 45 (angle at the circumference).
AB is a diameter, APB = 90 .
Let R be the point of intersection of AQ and BP. In APR, APR = 90 and PAR = 45 , so ARB = 45 (constant).
Hence R lies on a circle through A and B on which chord AB subtends 45 , i.e. 90 at its centre. Its centre (0, k) is on the perpendicular bisector of AB and its radius = AB 2 = 2 2, so
k2 + 4 = 8 k = 2 (R is above the x-axis since P is in the first quadrant)
locus: x2 + (y - 2)2 = 8 x2 + y2 - 4y - 4 = 0
Check: = 3 gives R(1 + 3, 1 + 3), which satisfies it.
Answer: (4)