ANSWER
Center: (2,-1)
Radius: 3 units.
EXPLANATION
The given circle has equation:
[tex] {x}^{2} + {y}^{2} - 4x + 2y - 4 = 0[/tex]
We rearrange to get;
[tex]{x}^{2} - 4x + {y}^{2}+ 2y =4[/tex]
We add the square of half the coefficients of the linear terms to both sides of the equation as shown below:
[tex]{x}^{2} - 4x + {( - 2)}^{2} + {y}^{2}+ 2y + {(1)}^{2} =4+ {( - 2)}^{2}+ {(1)}^{2} [/tex]
Look out for the perfect squares on the left hand side .
[tex] {(x - 2)}^{2} +{(y + 1)}^{2} =4+ 4+ 1[/tex]
[tex]{(x - 2)}^{2} +{(y + 1)}^{2} =9[/tex]
[tex]{(x - 2)}^{2} +{(y + 1)}^{2} = {3}^{2} [/tex]
By comparing to :
[tex]{(x - h)}^{2} +{(y - k)}^{2} = {r}^{2} [/tex]
We have the center to be
(h,k)=(2,-1) and radius r=3.
