A circle x^2 + y^ + 2x - 2y -1 = 0 expressed in the following form (x-h)^2 + (y-k)^2 = r^2 where h, k are the coordinates of the centre. How is r the radius given?
eh? r is the radius. r = √r^2
Not sure what you mean by "how is r given?" Its square is part of the equation, as are h and k.
For this particular circle, we have
x^2+2x + y^2-2y = 1
x^2+2x+1 + y^2-2y+1 = 1+1+1
(x+1)^2 + (y-1)^2 = 3
so, r = √3