$$S \equiv x^2+y^2-2 x-4 y-4=0$$ and $$S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0$$ are two circles the point $$(-2,-1)$$ lies

The centre of the circle passing through $$(0,0)$$ and $$(1,0)$$ and touching the circle $$x^2+y^2=9$$ is

If two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then

the circle $${x^2} + {y^2} + 4x - 7y + 12 = 0$$ cuts an intercept on Y-axis of length