If the equation of the circle which cuts each of the circles $x^2+y^2=4, x^2+y^2-6 x-8 y+10=0$ and $x^2+y^2+2 x-4 y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2 g x+2 f y+c=0$, then $g+f+c=$

Since the circle $x^2+y^2+2 g x+2 f y+c=0$ cuts each given circles at extremities of a diameter Then the common chords will pass through the centre of the respective circles. So, common chord of $x^2+y^2=4$ and $x^2+y^2+2 g x+$ $2 f y+c=0$ is $2 g x+2 f y+c+4=0$ passes through $(0,0)$ $\Rightarrow c=-4$ Next, common chord of $x^2+y^2-6 x-8 y+10=0$ and $x^2+y^2+2 g x+2 f y+c=0$ is $2 g x+2 f y+c+6 x+8 y-10$ $=0$ passes through $(3,4)$ $\Rightarrow 3 g+4 f+18=0\ldots(i)$ and, common chord of $x^2+y^2+2 x-4 y-2=0$ and $x^2+y^2+2 g x+2 f y+c=0$ is $2 g x+2 f y+c-2 x+4 y+2$ $=0$ passes through $(-1,2)$ $\Rightarrow g-2 f-4=0$
From (i) and (ii), $f=-3, g=-2$ Hence, $g+f+c=-9$