Determine the nature of the loci $\;$ $x^2 + y^2 + 4x + 6y + 16 = 0$
Equation of given locus is: $\;$ $x^2 + y^2 + 4x + 6y + 16 = 0$
Completing the square, we have,
$\left(x^2 + 4x + 4\right) + \left(y^2 + 6y + 9\right) - 4 - 9 + 16 = 0$
i.e. $\;$ $\left(x + 2\right)^2 + \left(y + 3\right)^2 + 3 = 0$
i.e. $\;$ $\left(x + 2\right)^2 + \left(y + 3\right)^2 = \sqrt{-3}$
Hence, the given equation represents an imaginary circle whose center is $\left(-2, -3\right)$ and
radius is $= \sqrt{-3} = i \sqrt{3}$ $\;$ where $\;$ $i = \sqrt{-1}$