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.