If the circle $x^2 + y^2 + 4x + 22y + c = 0$ bisects the circumference of the circle $x^2 + y^2 - 2x + 8y - d = 0$,then $c + d = . . . . .$

If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4y+2=0$ touch each other,then $\lambda=$

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

If two circles touch at $(0,0)$ externally,then their radical axis is

The radius of the circle whose centre lies at $(1, 2)$,while cutting the circle $x^2 + y^2 + 4x + 16y - 30 = 0$ orthogonally,is (in units):