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#.
