First; every circle can be written in the form $x^2+y^2+ux+vy+w=0$ and every such equation represents a circle. Thus $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ is a circle iff $k\ne -1$.

Also for the intersection point $(x,y)$ we have $x^2 + y^2 + Ax + By + C=0$ and $x^2 + y^2 + ax + by + c=0$ hence for such a point $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ so the intersection points of the two circles lie on the third circle.