The poles of the tangents to the circle $$x^2+y^2=4$$ with respect to the circle $$(x+2)^2+y^2=8$$, lie on

If the power of the point $$(1,6)$$ with respect to the circle $$x^2+y^2+4 x-6 y-a=0$$ is $$-16$$ then $$a$$ equals

The equation of radical axis of the circles $$x^2+y^2+4 x+6 y+7=0$$ and $$4 x^2+4 y^2+8 x+12 y-9=0$$ is

The radical axis of the circles $$S_1: x^2+y^2-4 x+6 y-10=0$$ and $$S_2 : x^2+y^2+2 x-6 y+2=0$$, cut the circle $$S_1$$ in