You are given an equation of a circle in the "general form".

The first step is to transform it into the "standard form".



For it, perform completing the squares for x- and y-terms separately.

I will do it below, step by step. Trace my steps (!)


    x^2 + y^2 + 6x - 4y = k^2 - k - 33.                       (<<<---=== the original equation)


    (x^2 + 6x) + (y^2 - 4y) = k^2 - k - 33                    (<<<---=== I grouped / re-grouped the terms)

    (x^2 + 6x + 9) + (y^2 - 4y + 4) = k^2 - k - 33 + 9 + 4    (<<<---=== I added some constant terms in both sides)

    (x+3)^2        + (y-2)^2        = k^2 - k - 20            (<<<---=== I completed the squares in the left side 
                                                                         and combined like terms in the right side)


Now I have a "standard form" of the circle equation.


In order for this equation describes a real circle, its right side should be positive.

If the right side is zero, then the circle becomes a single point.

If the right side is negative, then the equation describes empty set.



So, if  k^2 - k - 20 > 0, then you have a circle with the center at the point (-3,2) in a coordinate plane (x,y);

    if  k^2 - k - 20 = 0, then the circle degenerates to the single point (-3,2);

    if  k^2 - k - 20 = 0, then the equation describes the empty set.



The parabola y = k^2 - k - 20 = (k-5)*(k+4)  has the roots k= 5 and k= -4.

    It is positive  if  k < -4  ot  k > 5;

    it is 0         if k= -4 or k= 5;

    it is negative  if  -4 < k < 5.



Therefore, the given (original) equation 

    - represents a circle,       if  k < -4  ot  k > 5;

    - represent a single point,  if  k= -4 or k= 5;

    - represents the empty set,  if  -4 < k < 5.