Question

Let $A\left(z_1\right)$ be the point of intersection of curves $\arg (z-2+i)=\frac{3\pi }{4}$ and $\arg (z+\sqrt{3}i)=\frac{\pi }{3}.$ $B\left(z_2\right)$ is the point on $\arg (z+\sqrt{3}i)=\frac{\pi }{3}$ such that $|z_2-5|$ is minimum, and $C\left(z_3\right)$ is the centre of circle $|z-5|=3.$ If the area of triangle $ABC$ is $\sqrt{k}$ sq. units, then the value of $k$ is

Answer 1

$A\left(z_1\right)$ is the point of intersection of $\arg (z-2+i)=\frac{3\pi }{4}$ and $\arg (z+\sqrt{3}i)=\frac{\pi }{3}$
Now, $\arg (z-2+i)=\frac{3\pi }{4}$
$\Rightarrow x+y=1\dots \left(1\right)$
and $\arg (z+\sqrt{3}i)=\frac{\pi }{3}$
$\Rightarrow y-\sqrt{3}x=-\sqrt{3}\dots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right),$ we get
$z_1=(1,0)$

$B\left(z_2\right)$ is the point on $\arg (z+\sqrt{3}i)=\frac{\pi }{3}$ such that $|z_2-5|$ is minimum.
Let $|z_2-5|=r$
$\Rightarrow (x_2-5{)}^2+y_2^2=r^2\dots (3)$
For minimum value of $r,$ line $\left(2\right)$ should be tangent to circle $\left(3\right)$
$\Rightarrow r=\left|\frac{5\sqrt{3}-\sqrt{3}}{\sqrt{3+1}}\right|=2\sqrt{3}$
$C\left(z_3\right)$ is the centre of the circle $|z-5|=3$
$\Rightarrow z_3=(5,0)$


Now, $AB=\sqrt{A{C}^2-B{C}^2}=\sqrt{16-12}=2$
Area of $\Delta ABC=\frac{1}{2}\times AB\times BC$
$=\frac{1}{2}\cdot 2\cdot 2\sqrt{3}=2\sqrt{3}=\sqrt{12}$ sq. units.