If the circle $S=0$ cuts the circles $x^2+y^2-2x+6y=0$,$x^2+y^2-4x-2y+6=0$,and $x^2+y^2-12x+2y+3=0$ orthogonally,then the equation of the tangent at $(0,3)$ on $S=0$ is

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

If $P(\alpha, \beta)$ is the radical centre of the circles $S \equiv x^2+y^2+4x+7=0$,$S^{\prime} \equiv 2x^2+2y^2+3x+5y+9=0$ and $S^{\prime \prime} \equiv x^2+y^2+y=0$,then the length of the tangent drawn from $P$ to $S^{\prime}=0$ is

When do the two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ intersect each other orthogonally?

If the circle $x^2+y^2+2kx+4y-4=0$ has its centre in the $4^{\text{th}}$ quadrant and touches the circle $x^2+y^2+6x-2y+6=0$,then $k=$