Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2 g x+2 f y+c=0$ and $S \equiv x^2+y^2-6 x-4 y+4=0$ and the radius of the circle $S=0$ be 1 . The $g+f=$

The radius of the circle which cuts all the three circles $x^2+y^2-4 x-4 y+3=0, x^2+y^2+4 x-4 y+3=0$ and $x^2+y^2+4 x+4 y+3=0$ orthogonally is

From a point $A(0,3)$ on the circle $(x+2)^2+(y-3)^2=4$, a chord $A B$ is drawn and it is extended to a point $Q$ such that $A Q=2 A B$. Then, the locus of $Q$ is

If $m_1, m_2$ are the slopes of the tangents drawn from a point $(1,-3)$ to the circle $x^2+y^2-6 x+4 y+12=0$, then $9\left(m_1^2+m_2^2\right)=$