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

Suppose $A(2,3)$ and $B$ are the points of intersection of two circles. The points $P$ lying on one circle and $Q$ lying on the other circle are such that $BP$ and $BQ$ constitute the diameters of the circles. If the slopes of the radical axis and $PQ$ are $3/4$ and $a/b$ respectively,then the value of $3a+4b$ is

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

Two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ will intersect each other orthogonally,only when

If one of the diameters of the circle $x^2+y^2-10x+4y+13=0$ is a chord of another circle $C$,whose center is the point of intersection of the lines $2x+3y=12$ and $3x-2y=5$,then the radius of the circle $C$ is

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