Let a,b in R and `a^2+ b^2 ne 0 `
Suppose `S={z in C : z =(1)/(alpha +I bt ), t in R, t ne 0 } `where `I= sqrt(-1)` if z=x +iy and `ne S,` then (x,y) lies on

Options

A. the circle with radius `1/(2 alpha)` and centre `(1/(2a),(0))` for `a gt 0, b ne 0 `
B. the circle with radius `-1/2a` and centre `(-1/2a,0)` for a `lt 0, b ne 0`
C. the X -axis for `a ne 0, b=0`
D. the` Y-axis for a-0 , b ne 0 `

Answer

D

Explanation

Here `x+iy=(1)/(a+ibt)xx(a -ibt)/(a-ibt)`
`therefore x+ iy =(a-ibt)/(a^2+b^2t^2)`
Let `a ne 0, b ne 0 `
`therefore x =(a)/(a^2+b^2t^2)and y = (-bt )/(a^2+b^2t^2)`
`rArr y/x = (-bt)/(a)rArr t= (ay)/(bx)`
On puttting `x=(a)/(a^2+b^2 t^2)`we get
`x(a^2+b^2,(a^2y^2)/(b^2 x^2))=arArr a^2(x^2 + y^2)`
or `x^2+y^2 -x/a=0`
or `x^2+y^2 -x/a=0`
of `(x -1/(2a))^2+y^2 =(1)/(4a^2)`
`therefore` Option (a) is correct
For `a ne 0 and b=0`
`x+ iy = 1/a rArr x =1/a, y=0`
`rArr` z lies on X-axis.
`therefore` Option ( c) is correct
For a=0 and `b ne 0, x+ iy = (1)/(ibt)rArrx=0 , y=-1/(bt)`
`rArr` z lies on Y -axis
`therefore` Option (d) is correct