P and Q are any two points on the circle `x^2+y^2= 4` such that PQ is a diameter. If `alpha` and `beta` are the lengths of perpendiculars from `P` and `Q` on `x + y = 1` then the maximum value of `alphabeta` is

Options

A. `(1)/(2)`
B. `(7)/(2)`
C. 1
D. 2

Answer

B

Explanation

Let `P(2 cos theta,2 sin theta), Q(-2 cos theta, -2 sin theta)`
`:. alpha beta =(|2cos theta+2 sin theta-1||-2cos theta-2 sin theta-1|)/(2)`
`=(|4(cos theta+sin theta)^(2)-1|)/(2)`
`=(|3+4sin2 theta|)/(2)`
`le (7)/(2)`