Question

Let a,b ` in ` R and `a^(2) + b^(2) ne 0` . Suppose `S = { z in C: z = (1)/(a+ ibt),t in R, t ne 0}`, where `i= sqrt(-1)`. If `z = x + iy` and z in S, then (x,y) lies on
A. the circle with radius `(1)/(2a)` and centre `((1)/(2a),0)` for `a gt 0 be ne 0`
B. the circle with radius `-(1)/(2a)` and centre `(-(1)/(2) ,0) a lt 0, b ne 0`
C. the axis for `a ne 0, b =0`
D. the y-axis for `a = 0, bne 0`

Answer 1

Correct Answer - A::C::D
`z =(1)/(a+ibt)`
`rArr x +iy = (a-ibt)/(a^(2) + b^(2)t^(2))`
`rArr x = (a)/(a^(2) + b^(2)t^(2)),y = (-bt)/(a^(2) =b^(2)t^(2))`
Eliminating t, we get
`x^(20 + y^(2) = (x)/(a)`
`rArr (x-(1)/(2a))^(2) +y^(2) = ((1)/(2a))^(2)`
`therefore `Option (1) is correct.
(3),(4) can be verified by putting b = 0 and a=0 respectively.