Let $z = x + iy$, where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\dfrac {z + i}{z - i}$ is purely imaginary lie on

Let $z = x + iy$

$\therefore \dfrac {z + i}{(z - i)}=\dfrac{x+iy+i}{x+iy-i}$

$= \dfrac {(x + i(y + 1))}{(x - i(1 - y))}\times \dfrac {(x + i(1 - y))}{(x + i(1 - y))}$

$=\dfrac {x^{2} + (y^{2} - 1)+2ix}{x^{2} + (1 - y)^{2}}$

Since, $\dfrac{z+i}{z-i}$ should be purely imaginary

$\therefore Re\left (\dfrac {z + i}{z - i}\right ) = 0$

$\implies \dfrac {x^{2} + (y^{2} - 1)}{x^{2} + (1 - y)^{2}} = 0$

$\implies x^{2}+y^{2}-1=0$

$\implies x^{2} + y^{2} = 1$ which is the equation of a circle

Hence, $(x,y)$ lie on a circle.