If the equation of the transverse common tangent of the circles $x^2+y^2-4 x+6 y+4=0$ and $x^2+y^2+2 x-2 y-2=0$ is $a x+b y+c=0$, then $\frac{a}{c}=$

A circle $S \equiv x^2+y^2+2 g x+2 f y+6=0$ cuts another circle $x^2+y^2-6 x-6 y-6=0$ orthogonally. If the angle between the circles $S=0$ and $x^2+y^2+6 x+6 y+2=0$ is $60^{\circ}$, then the radius of the circle $S=0$ is

If $m_1$ and $m_2$ are the slopes of the direct common tangents drawn to the circles $x^2+y^2-2 x-8 y+8=0$ and $x^2+y^2-8 x+15=0$, then $m_1+m_2=$

If the circumcenter of the triangle formed by the points $A(a, 3), B(b, 5)$ and $C(a, b)$ is $(1,1)$, then out of all the possible coordinates of $C$ the sum of the absolute values of the distinct coordinates of $C$ is