The circumference of the circle $x^2 + y^2 - 2x + 8y - q = 0$ is bisected by the circle $x^2 + y^2 + 4x + 12y + p = 0$. Then $p + q$ is equal to

Find the equation of the circle which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$,$x^2+y^2+5x-5y+9=0$,and $x^2+y^2+7x-9y+29=0$.

The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8x-10y-8=0$ and $x^2+y^2+2x-2y-2=0$ is

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

Let the set of all values of $r$,for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points,be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

If the equation of the circle which passes through the point $(1,1)$ and cuts both the circles $x^2+y^2-4x-6y+4=0$ and $x^2+y^2+6x-4y+15=0$ orthogonally is $x^2+y^2+2gx+2fy+c=0$,then $5g+2f+c=$