The gradient of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ is

If $(\alpha, \beta)$ is the centre of the circle which passes through the point $(1, -1)$ and cuts the circles $x^2+y^2+2x-3y-5=0$ and $x^2+y^2-3x+2y+1=0$ orthogonally,then $\alpha-5\beta=$

$P$ is a point of intersection of the circles $S \equiv x^2+y^2-6x+2ky+1=0$ and $S' \equiv x^2+y^2+2kx-6y-7=0$. If the tangent at $P$ to $S=0$ passes through the centre of $S'=0$ and the tangent at $P$ to $S'=0$ passes through the centre of $S=0$,then the radius of $S'=0$ is

$A$ 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

The equation of the circle passing through the points of intersection of two circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ and the point $(-1,1)$ is

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to