The locus of the centres of the circles which touch externally the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 4ax = 0$ will be

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

$A$ circle $S$ given by $x^2+y^2-14x+6y+33=0$ cuts the $X$-axis at $A$ and $B$ $(OB > OA)$. $C$ is the midpoint of $AB$. $L$ is a line through $C$ with slope $-1$. If $L$ is the diameter of a circle $S^{\prime}$ and also the radical axis of the circles $S$ and $S^{\prime}$,then the equation of the circle $S^{\prime}$ is

If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$