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.

Given, $$z = {{(1 - t){z_1} + t\,{z_2}} \over {(1 - t) + t}}$$

Clearly, z divides z₁ and z₂ in the ratio of t : (1 $$-$$ t), 0 < t < 1

$$\Rightarrow$$ AP + BP = AB

i.e., $$|z - {z_1}| + |z - {z_2}| = |{z_1} - {z_2}|$$

$$\Rightarrow$$ Option (a) is true.

and $$\arg (z - {z_1}) = \arg ({z_2} - z) = \arg ({z_2} - {z_1})$$

$$\Rightarrow$$ (b) is false and (d) is true.

Also, $$\arg (z - {z_1}) = \arg ({z_2} - {z_1})$$

$$ \Rightarrow \arg \left( {{{z - {z_1}} \over {{z_2} - {z_1}}}} \right) = 0$$

$$\therefore$$ $${{z - {z_1}} \over {{z_2} - {z_1}}}$$ is purely real.

$$ \Rightarrow {{z - {z_1}} \over {{z_2} - {z_1}}} = {{\overline z - {{\overline z }_1}} \over {{{\overline z }_2} - {{\overline z }_1}}}$$

or, $$\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right| = 0$$

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

Hence, (a, c, d) is the correct option.