The equation of the circle passing through the points of intersection of the circles $x^2 + y^2 - 6x + 8 = 0$ and $x^2 + y^2 - 6 = 0$ and the point $(1, 1)$ is:

The combined equation of the direct common tangents of the circles $x^2+y^2+2x=0$ and $x^2+y^2-2y-3=0$ is

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A=(4,1)$,where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Midpoints of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB=\sqrt{5}$,then the value of $r^2$ is

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. If the radius of the circle $S=0$ is $1$,then find the value of $g+f$.