Question

Let $a,b\in R$ and $a^2+b^2\ne 0$. Suppose $S=\{z\in C:z=\frac{1}{a+ibt},t\in R,t\ne 0\}$, where $i=\sqrt{-1}$. If $z=x+iy$ and $z\in S$, then $(x,y)$ lies on

• A. The circle with radius $\frac{1}{2a}$ and centre $(\frac{1}{2a},0)$ for $a>0,b\ne 0$
• B. The circle with radius $\frac{1}{2a}$ and centre $(-\frac{1}{2a},0)$ for $a<0,b\ne 0$
• C. The x-axis for $a\ne 0,b=0$
• D. The y-axis for $a=0,b\ne 0$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The circle with radius $\frac{1}{2a}$ and centre $(\frac{1}{2a},0)$ for $a>0,b\ne 0$
C The x-axis for $a\ne 0,b=0$
D The y-axis for $a=0,b\ne 0$

$x+iy=\frac{a-ibt}{a^2+b^2{t}^2}$

$x=\frac{a}{a^2+b^2{t}^2}$ …(1)

$y=-\frac{bt}{a^2+b^2{t}^2}$ …(2)

If $a=0,b\ne 0,x=0$, then we get option D

If $a\ne 0,b=0,y=0$, then we get option C

$a^2+b^2{t}^2=\frac{a}{x}$ and $a^2+b^2{t}^2=\frac{-bt}{y}$

$\therefore \frac{a}{x}=\frac{-bt}{y}\Rightarrow t=\frac{-ay}{bx}$ ….(3)

Putting (3) in (1), we get

$x(a^2+b^2\frac{a^2{y}^2}{b^2{x}^2})=a$

$\Rightarrow x(a^2+\frac{a^2{y}^2}{x^2})=a$

$\Rightarrow a^2(x^2+y^2)=ax$

$\Rightarrow x^2+y^2-\frac{1}{a}x=0$

Circle with center $(\frac{1}{2a},0)$

Radius $=\sqrt{{\left(\frac{-1}{2a}\right)}^2+0^2-0}=\frac{1}{2a}$