If the circles $x^2+y^2-4x+2fy+1=0$ and $x^2+y^2+2gx-4y-1=0$ cut orthogonally,then $r_1^2+r_2^2-8=$

If the circles given by $S \equiv x^2+y^2-14x+6y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0$ $(a \in N)$ have $4$ common tangents,then the possible number of circles $S^{\prime}=0$ is

If $S \equiv 2x^2+2y^2-8x+8y-7=0$ is the circle passing through the points of intersection of the circles $x^2+y^2+kx-ky+1=0$ and $x^2+y^2-kx+ky-2=0$,then the length of the tangent drawn from the point $(k, k)$ to the circle $S$ is

Two circles $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut each other orthogonally,then:

The equation of the radical axis of the circles $2x^2 + 2y^2 - 7x = 0$ and $x^2 + y^2 - 4y - 7 = 0$ is

If $(h, k)$ is the centre of the circle which passes through the origin and cuts the circles $x^2+y^2+4x+6y+12=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally,then $k-2h=$