Question

Let one of the vertices of the equilateral triangle circumscribing the circle $|z-2\sqrt{3}i|=1$ is $z_1=1+3\sqrt{3}i$. If the other two vertices are represented by $z_2$ and $z_3$, then which of the following statement(s) is/are CORRECT?

• A. The area of triangle is $\frac{3\sqrt{3}}{4}$.
• B. $\left|\mathrm{Re}\left(z_2{z}_3\right)\right|+\left|I_m\left(z_2{z}_3\right)\right|=8\text{. }$
• C. The radius of circle circumscribing the triangle is 2 .
• D. The perimeter of the escribed circle drawn opposite to vertex $z_1$ of triangle is $\frac{3\pi }{2}$.

Answer 1

Answer: (B), (C)

By using rotation, we get $\frac{z-2\sqrt{3}i}{(1+3\sqrt{3}i)-2\sqrt{3}i}=e^{\pm \frac{i2\pi }{3}}$
$\Rightarrow z=2\sqrt{3}i+(1+i\sqrt{3})\left(\frac{-1}{2}\pm \frac{i\sqrt{3}}{2}\right)$
$\therefore z=-2+2\sqrt{3}i,1+i\sqrt{3}$
Clearly radius of circle circumscribing the triangle is $2\Rightarrow$ option $(C)$ is correct
Now $z_2{z}_3=(-2+2\sqrt{3}i)(1+\sqrt{3}i)=-2-6-2\sqrt{3}i+2\sqrt{3}i=-8+i0$

$\Rightarrow \left|\mathrm{Re}\left(z_2{z}_3\right)\right|+\left|I_m\left(z_2{z}_3\right)\right|=8$
$\Rightarrow$ Option $(B)$ is correct.
Also area $(△ABC)=\frac{\sqrt{3}}{4}(2\sqrt{3}{)}^2=3\sqrt{3}$
$\Rightarrow$ Option (A) is incorrect.
Now $r_1=\frac{\Delta }{s-a}=\frac{3\sqrt{3}}{\frac{6\sqrt{3}}{2}-2\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{3}}=3$
Hence the perimeter of the escribed circle $=6\pi \Rightarrow$ Option (D) is incorrect