Question

Given that the two curves arg(z) = $\frac{\pi }{6}$ and $|z-2\sqrt{3}i|=r$ intersect in two distinct points, then
(Note : $\left[r\right]$ represents integral part of $r$)

• A. $\left[r\right]\ne 2$
• B. $0<r<3$
• C. $r=6$
• D. $3<r<2\sqrt{3}$

Answer 1

Answer: (A), (D)

$\left[r\right]\ne 2$
$3<r<2\sqrt{3}$

$arg\left(z\right)=\frac{\pi }{6}|z-2\sqrt{3}i|=r$
Let, $z=x+iy$
$y=\frac{x}{\sqrt{3}}$ ...(1)
$x^2{+(y-2\sqrt{3})}^2=r^2$ ... (2)
Substitute equation (1) in equation (2), we get
$\Rightarrow {\left(y\sqrt{3}\right)}^2{+(y-2\sqrt{3})}^2=r^2\Rightarrow 4y^2-4\sqrt{3}y+12-r^2=0$
Since the straight line intersects the circle in two distinct points.
$\therefore b^2-4ac>0$ for the above quadratic
$\Rightarrow 48-16(12-r^2)>0\therefore r>3\&r\ne 2$
For $arg\left(z\right)=\frac{\pi }{6}$, $y>0$.
Hence, the product of the roots of $y$ must be positive.
Hence,
$12-r^2>0$. Hence, $r<2\sqrt{3}$
Hence, $3<r<2\sqrt{3}$.
Hence, options A and D are correct.