Determine the nature of the loci $\;$ $\left(x + 7\right) \left(x + 9\right) + \left(y - 1\right) \left(y + 3\right) = 11$
Equation of given locus is: $\;$ $\left(x + 7\right) \left(x + 9\right) + \left(y - 1\right) \left(y + 3\right) = 11$
i.e. $\;$ $x^2 + 16x + 63 + y^2 + 2y - 3 = 11$
i.e. $\;$ $x^2 + 16x + y^2 + 2y + 49 = 0$
Completing the square, we have,
$\left(x^2 + 16x + 64\right) + \left(y^2 + 2y + 1\right) + 49 - 64 - 1 = 0$
i.e. $\;$ $\left(x + 8\right)^2 + \left(y + 1\right)^2 - 16 = 0$
i.e. $\;$ $\left(x + 8\right)^2 + \left(y + 1\right)^2 = \left(4\right)^2$
Hence, the given equation represents a real circle whose center is $\left(-8, -1\right)$ and
radius is $= \sqrt{16} = 4$