Answer
center: $(-2, 1)$
radius = $3$ units
Refer to the image below for the graph.
Work Step by Step
Add 4 to both sides of the equation to obtain:
$x^2+4x+y^2-2y=4$
Group the terms with the same variables to obtain:
$(x^2+4x)+(y^2-2y)=4$
Complete the square for each group by adding $(\frac{4}{2})^2=2^2=4$ in the x-group and adding $(\frac{-2}{2})^2=(-1)^2=1$ to the y-group. Add 4 and 1 to the right side to maintain the equality of both sides.
$(x^2+4x+4) + (y^2-2y+1) = 4+4+ 1
\\(x^2+4x+4) + (y^2-2y+1) = 9$
Factor each trinomial to obtain:
$(x+2)^2 + (y-1)^2=9
\\(x+2)^2+(y-1)^2=3^2$
RECALL:
The standard form of a circle's equation is $(x-h)^2 + (y-k)^2=r^2$ where $(h, k)$ is its center and $r$ = radius.
Thus, the circle $(x+2)^2+(y-1)^2=3^2$ has its center at $(-2, 1)$ and its radius is $3$ units.
To graph the circle, perform the following steps;
(1) Plot the center $(-2, 1)$.
(2) With a radius of $3$, plot the points that are 3 units directly above, below, to the left, and to the right of the circle
(3) Connect the four points (not including the center of the circle) using a smooth curve to form a circle.
(Refer to the attached image in the answer part above for the graph.)
