The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

The centre of the circle passing through the point $(1,1)$ and orthogonal to the circles $x^2+y^2+3x-5y+7=0$ and $x^2+y^2-6x-10y+9=0$ is

The equation of the circle which cuts all the three circles $4(x-1)^2+4(y-1)^2=1$,$4(x+1)^2+4(y-1)^2=1$,and $4(x+1)^2+4(y+1)^2=1$ orthogonally is

If $x^2 + y^2 + px + 3y - 5 = 0$ and $x^2 + y^2 + 5x + py + 7 = 0$ cut orthogonally,then $p$ is

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

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