How do you write the general form given a circle that passes through the given points. P(-2, 0), Q(1, -3), R(-2, -6)?

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`