Question

Let $A\left(z_1\right)$ be the point of intersection of curves $\arg \left(z-2+i\right)=\frac{3\pi }{4}$ and $\arg \left(z+i\sqrt{3}\right)=\frac{\pi }{3}$. $B\left(z_2\right)$ be the point on the curve $\arg \left(z+i\sqrt{3}\right)=\frac{\pi }{3}$ such that $\left|z_2-5\right|$ is minimum and $C\left(z_3\right)$ be the centre of circle $\left|z-5\right|=3$.
[Note : $i^2=-1$]
The area of triangle ABC is equal to:

• A. $4\sqrt{3}$
• B. $\frac{3\sqrt{3}}{2}$
• C. $2\sqrt{3}$
• D. 4

Answer 1

The correct option is C $2\sqrt{3}$

$arg(z-2+i)=\frac{3\pi }{4}$

Let $z=x+iy$

$arg(x-2+iy+i)=\frac{3\pi }{4}$

$\frac{1+y}{x-2}=tan\left(\frac{3\pi }{4}\right)=-1$

$x+y=1-\left(I\right)$

$arg(z+i\sqrt{3})=\frac{\pi }{3}\Rightarrow arg(x+iy+i\sqrt{3})=\frac{\pi }{3}$

$\frac{y+\sqrt{3}}{x}=tan\frac{\pi }{3}=\sqrt{3}$

$y=x\sqrt{3}-\sqrt{3}-\left(II\right)$

Solving (I) and (II), we get

$x=1,y=0\Rightarrow (1,0)$

Now, $y=x\sqrt{3}-\sqrt{3}=\sqrt{3}(x-1)$

$|z_2-5|\Rightarrow$ Minimum

Let $z_2=x+iy$

$\sqrt{(x-5{)}^2+y^2}\Rightarrow$ Minimum = $z$ (arg)

$\frac{dz}{dx}=\frac{1}{2\sqrt{(x-5{)}^2+y^2}}\times \left(2\right(x-5)+2y\frac{dy}{dx})$

for Minima of $z,\frac{dz}{dx}=0$

Therefore, $\frac{dy}{dx}=-\frac{(x-5)}{y}=\frac{-(x-5)}{\sqrt{3}(x-1)}-\left(III\right)$

$y=x\sqrt{3}-\sqrt{3}$

$\frac{dy}{dx}=\sqrt{3}-\left(IV\right)$

From (III) and (IV),

$\sqrt{3}=\frac{-(x-5)}{\sqrt{3}(x-1)}$

$3x-3=-x+5$

$4x=8$

$x=2\Rightarrow y=\sqrt{3}\Rightarrow (2,\sqrt{3})$

Now $|z-5|=3$

This is equation of circle with centre $(5,0)$

Therefore, points are $A(1,0);B(2,\sqrt{3});C(5,0)$

Area $=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ 2& \sqrt{3}& 1\\ 5& 0& 1\end{array}\right|=2\sqrt{3}$