The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

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

$A$ circle $S$ cuts three circles $x^2+y^2-4x-2y+4=0$,$x^2+y^2-2x-4y+1=0$,and $x^2+y^2+4x+2y+1=0$ orthogonally. Then,the radius of $S$ 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

The minimum distance between any two points $P_{1}$ and $P_{2}$,where $P_{1}$ lies on the first circle and $P_{2}$ lies on the second circle,for the given equations:
$x^{2}+y^{2}-10x-10y+41=0$
$x^{2}+y^{2}-24x-10y+160=0$
is .........

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :