Let the point P represent z=x+iy, where x, y | vedclass

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(D) For condition $(i)$, $\frac{2z+i}{z-2} = \frac{2(x+iy)+i}{(x-2)+iy} = \frac{2x + i(2y+1)}{(x-2)+iy}$.For this to be purely imaginary, the real par

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$\left(\frac{5}{37},-\frac{30}{37}\right)$

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(D) For condition $(i)$, $\frac{2z+i}{z-2} = \frac{2(x+iy)+i}{(x-2)+iy} = \frac{2x + i(2y+1)}{(x-2)+iy}$.For this to be purely imaginary, the real part must be zero:$\operatorname{Re}\left(\frac{2x + i(2y+1)}{(x-2)+iy} \cdot \frac{(x-2)-iy}{(x-2)-iy}\right) = 0 \implies 2x(x-2) + y(2y+1) = 0$.$2x^2 - 4x + 2y^2 + y = 0 \implies x^2 + y^2 - 2x + \frac{1}{2}y = 0$ (Circle $C_1$).For condition $(ii)$, $\operatorname{Arg}\left(\frac{z+i}{z+1}\right) = \frac{\pi}{2}$ implies $\frac{z+i}{z+1}$ is purely imaginary with a positive imaginary part.Let $z+i = x+i(y+1)$ and $z+1 = (x+1)+iy$.$\frac{z+i}{z+1} = \frac{(x+i(y+1))((x+1)-iy)}{(x+1)^2+y^2} = \frac{x(x+1) + y(y+1) + i((x+1)(y+1) - xy)}{(x+1)^2+y^2}$.Real part $x(x+1) + y(y+1) = 0 \implies x^2 + x + y^2 + y = 0$ (Circle $C_2$).Subtracting the equations of $C_1$ and $C_2$:$(x^2 + y^2 - 2x + \frac{1}{2}y) - (x^2 + y^2 + x + y) = 0 \implies -3x - \frac{1}{2}y = 0 \implies y = -6x$.Substituting $y = -6x$ into $x^2 + y^2 + x + y = 0$:$x^2 + 36x^2 + x - 6x = 0 \implies 37x^2 - 5x = 0$.$x(37x - 5) = 0 \implies x = 0$ or $x = \frac{5}{37}$.For $x = \frac{5}{37}$, $y = -6(\frac{5}{37}) = -\frac{30}{37}$.The intersection point other than the origin is $\left(\frac{5}{37}, -\frac{30}{37}\right)$.

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