The equation of the family of circles passing through the point of intersection of a circle $S = 0$ and a line $P = 0$ is given by:

The number of common tangents that can be drawn to the circles $x^2+y^2-2x-2y-23=0$ and $x^2+y^2-4x-4y-1=0$ is

The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is

Let $a=1+i$ and $z=x+iy$. If the curve $z\bar{z}+az+\bar{a}\bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ is

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

If $x^2+y^2-6x-8y+12=0$ and $x^2+y^2-4x+6y+k=0$ cut orthogonally,then $k=$