The number of common tangents to the circles $x^2+y^2+4x-6y-12=0$ and $x^2+y^2-8x+10y+5=0$ is

Find the equation of the radical axis of the two circles $x^2 + y^2 - x + 1 = 0$ and $3(x^2 + y^2) + y - 1 = 0$.

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

If $S = x^2 + y^2 + 2x + 17y + 4 = 0$,$S' = x^2 + y^2 + 7x + 6y + 11 = 0$,and $S'' = x^2 + y^2 - x + 22y + 3 = 0$ are three circles,then the length of the tangent from their radical center to $S = 0$ is ......... units.

If the angle between the circles $x^2+y^2-2x+ky+1=0$ and $x^2+y^2-kx-2y+1=0$ is $\cos^{-1}(\frac{1}{4})$ and $k < 0$,then the point which lies on the radical axis of the given circles is