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

The equation of the circle which cuts the circles $S_1 \equiv x^2+y^2-4=0$,$S_2 \equiv x^2+y^2-6x-8y+10=0$,and $S_3 \equiv x^2+y^2+2x-4y-2=0$ at the extremities of the diameters of these circles is:

The centre of a circle which cuts $x^{2}+y^{2}+6x-1=0$,$x^{2}+y^{2}-3y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

If the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$ touch externally,then $\lambda$ 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