If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals

$\left(0, \frac{3}{4}\right)$ 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 $(g, f) =$

The equation of a circle,which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$,is

The equation of the tangent at the point $(0, 3)$ on the circle which cuts the circles $x^2 + y^2 - 2 x + 6 y = 0$,$x^2 + y^2 - 4 x - 2 y + 6 = 0$ and $x^2 + y^2 - 12 x + 2 y + 3 = 0$ orthogonally is

The radical centre of the circles $x^2+y^2+2x+3y+1=0$,$x^2+y^2+x-y+3=0$,and $x^2+y^2-3x+2y+5=0$ is

The length of the diameter of the circle which cuts the following three circles orthogonally is:
$x^{2}+y^{2}-x-y-14=0$
$x^{2}+y^{2}+3x-5y-10=0$
$x^{2}+y^{2}-2x+3y-27=0$