let a, b R and a²+b² ≠ 0.Suppose S=\{z C : z=1/a+ibt, t R, t ≠ 0\}, where i={-1} If z = x + iy and z S, then (x,y) lies on [JEE Adv 2016 P2]

the circle with radius $\frac{1}{2a}$ and center $\left(\frac{1}{2a},0\right)$ for $a>0, b \neq 0$
the circle with radius $\frac{1}{2a}$ and center $\left(\frac{1}{2a},0\right)$ for $a<0, b \neq 0$
the x - axis for $a \neq 0, b=0$
the y - axis for $a=0, b \neq 0$

Solution:

x+iy=\frac{1}{a+ibt}

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

Let

a \neq 0 \; \& \; b \neq 0

x=\frac{a}{a^2+b^2t^2}

...............(1)

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

...............(2)

\frac{y}{x}=\frac{-bt}{a} \Rightarrow t=-\frac{ay}{bx}

put in (1)

x\left(a^2+b^2\cdot\frac{a^2y^2}{b^2x^2}\right)=a

x^2+y^2-\frac{1}{ax}=0

\left(x-\frac{1}{2a}\right)^2+y^2=\frac{1}{4a^2}

⇒ option A is correct

for a ≠ 0,b = 0

x+iy=\frac{1}{a}

x=\frac{1}{a}, y=0

⇒z lies on x-axis ⇒ option (C) i correct

for a = 0,b ≠ 0

x+iy=\frac{1}{ibt}

y=-\frac{1}{bti}, x=0

⇒ z lies on y-axis. ⇒ option (D) is correct

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