Question

Let the point $P$ represent $z=x+iy,a,x,y\in R$ in the Argand plane. Let the curves $C_1$ and $C_2$ be the loci of P satisfying the conditions
(i) $\frac{2z+i}{z-2}$ is purely imaginary and
(ii) $Arg\left(\frac{z+i}{z+1}\right)=\frac{\pi }{2}$ respectively. Then the point of intersection of the curves $C_1$ and $C_2$, other than the origin, is

• A. (1, 2)
• B. $\left(\frac{2}{7},-\frac{5}{7}\right)$
• C. (-3. 4)
• D. $\left(\frac{5}{37},-\frac{30}{37}\right)$

Answer 1

Answer: (C)

If $z=x+iy,x,y\in R$
then $\frac{2z+i}{z-2}=\frac{2(x+iy)+i}{(x+iy)-2}$
$=\frac{2x+i(2y+1)}{(x-2)+iy}$
$=\frac{2x+i(2y+1)}{(x-2)+iy}\times \frac{(x-2)-iy}{(x-2)-iy}$
$\because Re\left(\frac{2z+i}{z-2}\right)=0$
$\Rightarrow 2x(x-2)+y(2y+1)=0$
$\Rightarrow 2x^2+2y^2-4x+y=0$
$\Rightarrow x^2+y^2-2x+\frac{y}{2}=0\dots$(i)
and $\frac{z+i}{z+1}=\frac{x+i(y+1)}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$
$\therefore \arg \left(\frac{z+i}{z+1}\right)$
$={\tan }^{-1}\left[\frac{(x+1)(y+1)-xy}{x(x+1)+y(y+1)}\right]=\frac{\pi }{2}$
$\Rightarrow x^2+y^2+x+y=0\dots$(ii)
Equation of common chord of circles (i) and (ii) is
$S_1-S_2=0$
$\Rightarrow -3x-\frac{y}{2}=0$
$\Rightarrow y+6x=0\dots$(iii)
For the intersection of the circles on solving chord
(iii) and circle (ii), we get
$x^2+36x^2+x-6x=0$
$\Rightarrow 37x^2-5x=0$
$\Rightarrow x=0,\frac{5}{37}$
So, $y$ -coordinate is $0,-\frac{30}{37}$.
$\therefore$ The point of intersection of the curves $C_1$ and $C_2$
other than the origin is $\left(\frac{5}{37},-\frac{30}{37}\right)$.