Question

Let $C_1$ and $C_2$ be two curves on the complex plane defined as $C_1:z+\overline{z}=2|z-1|$ and $C_2:\arg (z+1+i)=\alpha ,$ where $\alpha \in (0,\frac{\pi }{2})$ such that curves $C_1$ and $C_2$ touch each other exactly at one point at $P\left(z_0\right).$ A particle starts from the point $P\left(z_0\right).$ It moves horizontally away from the origin by $2$ units and then vertically away from the origin by $3$ units to reach at a point $P\left(z_1\right).$ Let $A$ denote the area bounded by $C_1$ and the line joining $P\left(z_0\right)$ and $Q\left(z_0^{\prime }\right),$ where $Q\left(z_0^{\prime }\right)$ is the new position of $P\left(z_0\right)$ when $P\left(z_0\right)$ is rotated about the origin through an angle of $2\alpha$ in clockwise direction. Then

• A. $|z_0|=\sqrt{2}$
• B. $|z_0|=2\sqrt{2}$
• C. $|z_1|=5$
• D. $A=\frac{2}{3}$ sq. units

Answer 1

Answer: (A), (C), (D)

The correct options are
A $|z_0|=\sqrt{2}$
C $|z_1|=5$
D $A=\frac{2}{3}$ sq. units

$C_1:z+\overline{z}=2|z-1|$
$\Rightarrow 2x=2|x-1+iy|$
$\Rightarrow x^2=(x-1{)}^2+y^2$
$\Rightarrow y^2=2x-1\Rightarrow y^2=2(x-\frac{1}{2})$

$C_2:\arg (z-(-1-i\left)\right)=\alpha$
Curve $C_2$ is a ray emanating from $(-1,-1)$ and making an angle $\alpha$ from the positive real axis.
Since $C_1$ and $C_2$ have exactly one common point,
$\therefore C_2$ must be a tangent to $C_1$.
$C_2:y+1=m(x+1)$

Solving $C_1$ and $C_2,$ we get
$y^2=2(\frac{y+1}{m}-1-\frac{1}{2})$
$\Rightarrow m{y}^2-2y+3m-2=0$
Put $D=0$
$\Rightarrow 4-4m(3m-2)=0$
$\Rightarrow 3m^2-2m-1=0\Rightarrow (3m+1)(m-1)=0\Rightarrow (3m+1)(m-1)=0\Rightarrow m=-\frac{1}{3},1$
As $\alpha \in (0,\frac{\pi }{2}),$
$\Rightarrow m=1$

$\therefore C_2:y+1=x+1\Rightarrow y=x$
Putting $y=x$ in the curve $C_1,$
$x^2=2x-1\Rightarrow (x-1{)}^2=0\Rightarrow x=1$
$\Rightarrow P\left(z_0\right)\equiv (1,1)$
$\Rightarrow |z_0|=\sqrt{2}$


From above graph, $P\left(z_1\right)\equiv (3,4)$
$\Rightarrow |z_1|=5$



Area of the shaded region,
$A=2\underset{1/2}{\overset{1}{\int }}\sqrt{2x-1}dx$
$=2{\left[\frac{(2x-1{)}^{3/2}}{\frac{3}{2}\times 2}\right]}_{1/2}^1$
$=\frac{2}{3}[1-0]=\frac{2}{3}$ sq. units