If a diameter of the circle $x^2+y^2-4x+6y-12=0$ is a chord of a circle $S$ whose centre is at $(-3, 2)$,then the radius of $S$ is

Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^2+y^2=36$ is one of those circles,then the other circle is

The value of $k$ such that the circles $x^2 + y^2 + kx + 4y + 2 = 0$ and $2(x^2 + y^2) - 4x - 3y + k = 0$ cut orthogonally 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 .........

$A$ circle $C$ passes through $(2a, 0)$ and the line $2x = a$ is the radical axis of the circle $C$ and the circle $x^2 + y^2 = a^2$. Then,

If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is: