The two circles ${x^2} + {y^2} - 4y = 0$ and ${x^2} + {y^2} - 8y = 0$

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

Two circles with equal radii intersecting at the points $(0, 1)$ and $(0, -1).$ The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is

The point $(2, 3)$ is a limiting point of a coaxial system of circles of which ${x^2} + {y^2} = 9$ is a member. The co-ordinates of the other limiting point is given by

Circles ${x^2} + {y^2} - 2x - 4y = 0$ and ${x^2} + {y^2} - 8y - 4 = 0$

The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that