The circle passing through the intersection of the circles $x^{2}+y^{2}-6x=0$ and $x^{2}+y^{2}-4y=0$,having its centre on the line $2x-3y+12=0$,also passes through the point:

If the circle $S \equiv x^2+y^2+2gx+4y+1=0$ bisects the circumference of the circle $x^2+y^2-2x-3=0$,then the radius of circle $S=0$ 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 .........

The distance between the centres of similitude of the circles $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$ is

$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2+4x-6y+12=0$. If the radius of the circle having $C_1C_2$ as its diameter is $r$,then $\frac{9}{2}r=$

If the circles $x^2+y^2-2 \lambda x-2 y-7=0$ and $3(x^2+y^2)-8 x+29 y=0$ are orthogonal,then $\lambda=$