If the origin lies on a diameter of the circle $x^2+y^2-4 x-2 y-4=0$, then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

If $\alpha \neq-4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4 x+8 y+6=0$, then the values of the $y$-intercept of the chord lie in the interval

$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2 x+4 y+1=0$ and $x^2+y^2+4 x-6 y+12=0$. If the radius of the circle having $C_1 C_2$ as its diameters is $r$, then $\frac{9}{2} r=$

Suppose the circle $S: x^2+y^2+2 g x+2 f y+c=0$ cuts orthogonally the two circles $S^{\prime}: x^2+y^2-4 x-6 y+11=0$ and $S^{\prime \prime}: x^2+y^2-10 x-4 y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes, then $2 g+2 f+c=$