The radical centre of the circles $x^2+y^2+3x+2y+1=0$,$x^2+y^2-x+6y+5=0$ and $x^2+y^2+5x-8y+15=0$ is

Consider the three circles: $S_{1} \equiv x^{2}+y^{2}-6x-6y+4=0$,$S_{2} \equiv x^{2}+y^{2}-2x-4y+3=0$,and $S_{3} \equiv x^{2}+y^{2}+2kx+2y+1=0$. If the radical centre of these three circles exists,then which of the following cannot be the value of $k$?

If the equation of the circle passing through the points of intersection of the circles $S_1: x^2 - 2x + y^2 - 4y - 4 = 0$ and $S_2: x^2 + 2x + y^2 + 4y - 4 = 0$ passes through the point $(3, 3)$,and its equation is $x^2 + y^2 + \alpha x + \beta y + \gamma = 0$,then find the value of $3(\alpha + \beta + \gamma)$.

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

$A$ circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$