If P is a point on the circle x2 + y2 = 4, Q is a point on the straight line 5x + y + 2 = 0 and x - y + 1 = 0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is .
Answer
2
Solution
∵ x - y + 1 = 0 is the perpendicular bisector of PQ, Q is the image of P(h, k) in this line.
x - h1 = y - k-1 = -2(h - k + 1)1 + 1
Q = (k - 1, h + 1)
Q lies on 5x + y + 2 = 0:
5(k - 1) + (h + 1) + 2 = 0 h = 2 - 5k
P lies on the circle: (2 - 5k)2 + k2 = 4
26k2 - 20k = 0 k = 0 or k = 1013
h = 2 or h = 2 - 5013 = - 2413
Sum of abscissae = 2 - 2413 = 213
13 213 = 2
Answer: 2