Here, $$x + iy = {1 \over {a + ibt}} \times {{a - ibt} \over {a - ibt}}$$

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

Let a $$\ne$$ 0, b $$\ne$$ 0

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

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

On putting $$x = {a \over {{a^2} + {b^2}{t^2}}}$$, we get

$$x\left( {{a^2} + {b^2}\,.\,{{{a^2}{y^2}} \over {{b^2}{x^2}}}} \right) = a$$

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

or, $${x^2} + {y^2} - {x \over a} = 0$$ ...... (i)

or, $${\left( {x - {1 \over {2a}}} \right)^2} + {y^2} = {1 \over {4{a^2}}}$$

$$\therefore$$ Option (a) is correct.

For a $$\ne$$ 0 and b = 0,

$$x + iy = {1 \over a} \Rightarrow x = {1 \over a},\,y = 0$$

$$\Rightarrow$$ z lies on X-axis.

$$\therefore$$ Option (c) is correct.

For a = 0 and b $$\ne$$ 0,

$$x + iy = {1 \over {ibt}}$$

$$ \Rightarrow x = 0,\,y = - {1 \over {bt}}$$

$$\Rightarrow$$ z lies on Y-axis.

$$\therefore$$ Option (d) is correct.