The standard equation for a circle is:
(x-h)^2+(y-k)^2=r^2
Where (h,k) is the center and r is the radius.
We want to get x^2+6x+y^2-2y+6=0 into that format so we can identity the center and radius. To do so, we need to complete the square on the x and y terms separately. Starting with x:
(x^2+6x)+y^2-2y+6=0
(x^2+6x+9)+y^2-2y+6=9
(x+3)^2+y^2-2y+6=9
Now we can go ahead and subtract 6 from both sides:
(x+3)^2+y^2-2y=3
We are left to complete the square on the y terms:
(x+3)^2+(y^2-2y)=3
(x+3)^2+(y^2-2y+1)=3+1
(x+3)^2+(y-1)^2=4
The equation of this circle is therefore (x+3)^2+(y-1)^2=4. Note this can be rewritten as (x-(-3))^2+(y-(1))^2=4, so the center (h,k) is (-3,1). The radius is found by taking the square root of the number on the right side of the equation (which, in this case, is 4). Doing so yields a radius of 2.
