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 $z=x+i y, x, y \in R$
then $\frac{2 z+i}{z-2}=\frac{2(x+i y)+i}{(x+i y)-2}$
$=\frac{2 x+i(2 y+1)}{(x-2)+i y}$
$=\frac{2 x+i(2 y+1)}{(x-2)+i y} \times \frac{(x-2)-i y}{(x-2)-i y}$
$\because Re\left(\frac{2 z+i}{z-2}\right)=0$
$\Rightarrow 2 x(x-2)+y(2 y+1)=0$
$\Rightarrow 2 x^{2}+2 y^{2}-4 x+y=0$
$\Rightarrow x^{2}+y^{2}-2 x+\frac{y}{2}=0 \dots$(i)
and $\frac{z+i}{z+1}=\frac{x+i(y+1)}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$
$\therefore \arg \left(\frac{z+i}{z+1}\right) $
$=\tan ^{-1}\left[\frac{(x+1)(y+1)-x y}{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
$S_{1}-S_{2}=0$
$\Rightarrow -3 x-\frac{y}{2}=0$
$\Rightarrow y+6 x=0 \dots$(iii)
For the intersection of the circles on solving chord
(iii) and circle (ii), we get
$ x^{2}+36 x^{2}+x-6 x =0 $
$ \Rightarrow 37 x^{2}-5 x =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 $\left(\frac{5}{37},-\frac{30}{37}\right)$.