Let `0 ne a, 0 ne b in R`. 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

Options

A. on the circle with radius `1/(2a)` and center `(-1/(2a),0)`
B. on the circle with radius `1/(2a)` and center `(1/(2a),0)`
C. on the `x`-axis
D. on the y-axis.

Answer

B

Explanation

We have, `z=1/(a+ibt)`
`rArr x+iy=1/(a+ibt)`
`rArr x+iy=(a-ibt)/(a^(2)+b^(2)t^(2))`
`rArr x=a/(a^(2)+b^(2)t^(2))`
`rArr y/x=-b/a rArr t=(-ay)/(bx)`
Eliminating t between `x=a/(a^(2)+b^(2)t^(2))` and `t=-(ay)/(bx)`, we get
`x=a/(a^(2)+(a^(2)y^(2))/x^(2)) rArr a^(2)x^(2)+a^(2)y^(2)=ax rArr x^(2)+y^(2)-1/ax=0`
`rArr y/x=-b/at rArr t=-(ay)/(bx)`
Eliminating t between `x=a/(a^(2)+b^(2)t^(2))` and `t=-(ay)/(bx)`, we get
`x=a/(a^(2)+(a^(2)y^(2))/(x^(2)) rArr a^(2)x^(2)+a^(2)y^(2)=ax rArr x^(2)+y^(2)-1/ax=0`
`rArr (x-1/(2a))^(2) + (y-0)^(2)=(1/(2a))^(2)`
Clearly, it is a circle of radius `1/(2a)` and center `(1/(2a),0)`.