We are given three points #P(-2,0)#, #Q(1,-3)# and #R(-2,-6)#.

The perpendicular bisectors of any two lines joining these points will intersect at the center of the circle passing through these points.

Slope of line joining PQ is #(-3-0)/(1-(-2))=-3/3=-1# and mid point of PQ is #((-2+1)/2,(0-3)/2)# or #(-1/2,-3/2)#. Hence its perpendicular bisector will pass through #(-1/2,-3/2)# and have a slope of #-1/-1=1#.

Hence equation of perpendicular bisector of PQ is #y-(-3/2)=1xx(x-(-1/2))# or #y+3/2=x+1/2# or #x=y+1#.

Slope of line joining QR is #(-6-(-3))/(-2-1)=-3/-3=1# and mid point of QR is #((-2+1)/2,(-6-3)/2)# or #(-1/2,-9/2)#. Hence its perpendicular bisector will pass through #(-1/2,-9/2)# and have a slope of #-1/1=-1#.

Hence equation of perpendicular bisector of PQ is #y-(-9/2)=-1xx(x-(-1/2))# or #y+9/2=-x-1/2# or #x+y+5=0#.

Solving #x=y+1# and #x+y+5=0# gives us #(-2,-3)#, which is center #O# and radius will be its distance with say #P(-2,0)#, which is #3#

Hence equation of circle will be #(x-(-2))^2+(y-(-3))^2=3^2#

or #(x+2)^2+(y+3)^2=9# or #x^2+4x+4+y^2+6y+9=9# and simplifying it general for of equation of circle would be #x^2+y^2+4x+6y+4=0#