If $(p, q)$ is the centre of the circle which cuts the three circles $x^2+y^2-2x-4y+4=0$,$x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y-11=0$ orthogonally,then $p+q=$

The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ 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

From any point on the circle $x^2 + y^2 = a^2$,tangents are drawn to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$. The angle between them is:

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

Find the value of $m+n$,if the circumference of the circle $x^2+y^2+8x+8y-m=0$ is bisected by the circle $x^2+y^2-2x+4y+n=0$.