If $(\alpha, \beta)$ is the external centre of similitude of the circles $x^2+y^2=3$ and $x^2+y^2-2x+4y+4=0$,then $\frac{\beta}{\alpha}=$

If $C_1$ and $C_2$ are the centres of similitude with respect to the circles $x^2+y^2-14 x+6 y+33=0$ and $x^2+y^2+30 x-2 y+1=0$,then the equation of the circle with $C_1 C_2$ as diameter is

$P$ is a point of intersection of the circles $S \equiv x^2+y^2-6x+2ky+1=0$ and $S' \equiv x^2+y^2+2kx-6y-7=0$. If the tangent at $P$ to $S=0$ passes through the centre of $S'=0$ and the tangent at $P$ to $S'=0$ passes through the centre of $S=0$,then the radius of $S'=0$ is

The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

If the circle $x^2+y^2+2kx+4y-4=0$ has its centre in the $4^{\text{th}}$ quadrant and touches the circle $x^2+y^2+6x-2y+6=0$,then $k=$

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2x+6y=0$,$S_2: x^2+y^2+2gx-2y+6=0$,and $S_3: x^2+y^2-12x+2fy+3=0$. If $S_2$ and $S_3$ intersect orthogonally,then $(g, f) =$