If the circles $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally,then $k=$

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $x^2+y^2=9$ is

If the circle $x^2 + y^2 + 2x - 4y - k = 0$ lies exactly between the circles $x^2 + y^2 + 2x - 4y - 4 = 0$ and $x^2 + y^2 + 2x - 4y - 20 = 0$,then $k = \dots$

$A$ 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 $(-8,1)$

If the circles $x^2+y^2+2 \alpha x+2 y-8=0$ and $x^2+y^2-2 x+\alpha y-14=0$ intersect orthogonally,then the distance between their centres is