The circles $x^2 + y^2 + 4x + 6y + 3 = 0$ and $2(x^2 + y^2) + 6x + 4y + C = 0$ will cut orthogonally,if $C$ equals

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 $g+f$.

Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

The equation of a circle,which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$,is

The line $Ax + By + C = 0$ cuts the circle $x^2 + y^2 + ax + by + c = 0$ at points $P$ and $Q$,and the line $A'x + B'y + C' = 0$ cuts the circle $x^2 + y^2 + a'x + b'y + c' = 0$ at points $R$ and $S$. If the four points $P, Q, R,$ and $S$ are concyclic,then $D = \left| {\begin{array}{*{20}{c}}{a - a'}&{b - b'}&{c - c'}\\A&B&C\\{A'}&{B'}&{C'}\end{array}} \right| = $