Question

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

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

Answer 1

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

$z=x+i y=\frac{1}{a+i b t}=\frac{a-i b t}{a^{2}+b^{2} t^{2}} $
$x=\frac{a}{a^{2}+b^{2} t^{2}}, y=\frac{-b t}{a^{2}+b^{2} t^{2}}$
$\therefore \frac{y}{x}=\frac{-b t}{a} $
$\Rightarrow t=\frac{-a y}{b x}$
$S o x\left(a^{2}+b^{2} \frac{a^{2} y^{2}}{b^{2} x^{2}}\right)=a$
$a^{2} x^{2}+a^{2} y^{2}=a x $
$x^{2}+y^{2}=\frac{x}{a} $
$\Rightarrow \left(x-\frac{1}{2 a}\right)^{2}+y^{2}=\frac{1}{4 a^{2}}$
circle with centre $\left(\frac{1}{2 a }, 0\right)$ and radius $=\frac{1}{2 a }$
for $a >0, b \neq 0 $
If $ b =0, a \neq 0 $
$ x + iy =\frac{1}{ a } $
$\Rightarrow x =\frac{1}{ a }, \,\,\,y =0$
So $x$-axis
If $ a=0, b \neq 0 $
$x+i y=\frac{1}{i b t} $
$\Rightarrow x=0 ; y=-\frac{1}{b t}$
So y-axis