It is given that
$z=x+iy, x, y\in \mathrm{ℝ}$

And $\frac{2z+i}{z-2}=\frac{2(x+iy)+i}{(x-2)+iy}$

$Re\left(\frac{2z+i}{z-2}\right)=\frac{2x+i(2y+1)}{(x-2)+iy}\times \frac{(x-2)-iy}{(x-2)-iy}$

$0=2 x(x-2)+y(2 y+1)=0$

$0=x^2+y^2-2x+\frac{y}{2} ...\left(\mathrm{i}\right)$

And, $\frac{z+i}{z-i}=\frac{x+i(y+1)}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$

Now, $\arg \left(\frac{z+i}{z-i}\right)={\tan }^{-1}\left[\frac{(x+1)(y+1)-xy}{x(x+1)+y(y+1)}\right]=\frac{\pi }{2}$

$x^2+y^{2}x+x+y=0 ...\left(\mathrm{ii}\right)$

Equation of common chord of circles $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ is
$S_1-S_2=0$
$-3x-\frac{y}{2}=0$

$y+6x=0 ...\left(\mathrm{iii}\right)$

From Equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{iii}\right),$ the intersection of the circles is, $x^2+36x^2+x-6x=0$

$x=0, \frac{5}{37}$
So the values of $y$ are $0, -\frac{30}{37}$
This implies the point of intersection of the curves $C_1$ and $C_2$ other than the origin is $\left(\frac{5}{37},-\frac{30}{37}\right)$