If the circles of same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally,then $a =$

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 centre of the circle,which cuts orthogonally each of the three circles $x^2 + y^2 + 2x + 17y + 4 = 0$,$x^2 + y^2 + 7x + 6y + 11 = 0$,and $x^2 + y^2 - x + 22y + 3 = 0$,is

The equation of the radical axis of the circles $x^2+y^2+4x+6y+7=0$ and $4x^2+4y^2+8x+12y-9=0$ is:

If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x - 4)^2 + (y - 7)^2 = 36$ intersect at two distinct points,then

The equation of the circle passing through $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$ is: