The equation $z \bar{z} + (2 - 3i) z + (2 + 3i) \bar{z} + 4 = 0$ represents a circle of radius (in $\text{ units}$)

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

The complex numbers $z_1, z_2$ and $z_3$ satisfying $\frac{z_1 - z_3}{z_2 - z_3} = \frac{1 - i\sqrt{3}}{2}$ are the vertices of a triangle which is

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\ldots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$,where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P: |z_2-z_1|+|z_3-z_2|+\ldots+|z_{10}-z_9|+|z_1-z_{10}| \leq 2 \pi$
$Q: |z_2^2-z_1^2|+|z_3^2-z_2^2|+\ldots+|z_{10}^2-z_9^2|+|z_1^2-z_{10}^2| \leq 4 \pi$
Then,

If $P(x, y)$ represents the complex number $z = x + i y$ in the Argand plane and $\operatorname{Arg} \left( \frac{z - 3 i}{z + 4} \right) = \frac{\pi}{2}$,then the equation of the locus of $P$ is

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$,if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$,then $|\alpha - \beta|$ is equal to $.............$.