If the equation $2 x-3 y+3=0,2 x+y+1=0$ and $6 x+4 y+1=0$ represent the sides of a triangle, then the equation of the circle passing through the vertices of this triangle is

If $T_1 T^{\prime}{ }_1$ and $T_2 T_2^{\prime}$ are the common tangents of the circles $S \equiv x^2+y^2-2 x-4 y-4=0$ and $S \equiv x^2+y^2+4 x+4=0$, where $T_1, T^{\prime}{ }_1, T_2, T^{\prime}{ }_2$ are the points of contact, then the distance between $T_1$ and $T_1^{\prime}$ is

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

If the circle $S \equiv x^2+y^2+2 g x+2 f y+c=0$ cuts each of the three circles $x^2+y^2+4 x+4 y+7=0$, $x^2+y^2-4 x+4 y+7=0$ and $x^2+y^2-4 x-4 y+7=0$ orthogonally, then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S=0$ is