Question 10828
The goal here is to get the equation of a circle in the standard form:

{{{(x-h)^2 + (y-k)^2 = r^2}}} This is the equation of a circle with radius r and center at (h, k).

The technique is the "complete the square" in x and y.

{{{x^2 + 6x + y^2 - 8y - 11 = 0}}}  Isolate the x and y terms by adding 11 to both sides.

{{{(x^2 + 6x) + (y^2 - 8y) = 11}}} Now complete the squares in x and y.  This process entails adding a constant term to the x-group and to the y-group so that when you factor the x-group and the y-group, you will have a binomial squared in x and a binomial squared in y.
The constant for the x-group is found by squaring one-half of the coefficient of the x-term. That's (6/2)^2 = 9.  Similarly for the y-group, you square one-half of the coefficient of the y-term. That's (-8/2)^2 = 16.

{{{(x^2 + 6x + 9) + (y^2 - 8y + 16) = 11 + 9 + 16}}} Don't forget to add the same numbers to both sides of the equation.

{{{(x^2 + 6x + 9) + (y^2 - 8y + 16) = 36}}} Now factor.

{{{(x + 3)^2 + (y - 4)^2 = 6^2}}} Now it looks like the standard form of the equation for a circle with radius of 6 and center at (-3, 4).