If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

The equation of the circle passing through the point of intersection of the circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and the point $(3, -3)$ is:

If the circles $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_2$ is $l$,then $\frac{l}{n^2} =$

The centre of the circle touching the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ at their point of contact and passing through the point $(1, -1)$ is

From which point are the lengths of the tangents drawn to the circles $x^{2} + y^{2} - 8x + 40 = 0$,$5x^{2} + 5y^{2} - 25x + 80 = 0$,and $x^{2} + y^{2} - 8x + 16y + 160 = 0$ equal?

The coordinates of the radical centre of the three circles $x^2 + y^2 - 4x - 2y + 6 = 0$,$x^2 + y^2 - 2x - 4y - 1 = 0$,and $x^2 + y^2 - 12x + 2y + 30 = 0$ are