Circle

Given,$s \equiv x^2 + y^2 + 2gx + 2fy + c = 0$And$s' \equiv x^2 + y^2 - 6x + 6y + 2 = 0$$\Rightarrow (x^2-6x)+(y^2+6y) = -2$$\Rightarrow (x^2-6x+9)+(y^2+6y+9)$$= -2+9+9$$\Rightarrow (x-3)^2+(y+3)^2=16$Thus, the center of circle$S'$is$(3,-3)$and radius,$r'=4$Now,$S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$$\Rightarrow (x+g)^2+(y+f)^2 = g^2+f^2-c$So, the center of circle$S$is$(-g,-f)$and radius,$r = \sqrt{g^2+f^2-c}$Since, radius of circle $S$ is 1 .$\therefore \sqrt{g^2+f^2-c} = 1$$\Rightarrow g^2+f^2-c = 1$Also, two circle touch externally,So, distance between $(-g,-f)$ and$(3,-3) = r + r'$$\Rightarrow \sqrt{(-g-3)^2+(-f+3)^2} = 4+1$$\Rightarrow (g+3)^2+(f-3)^2=25$$\Rightarrow g^2+6g+9+f^2-6f+9=25$$\Rightarrow g^2+f^2+6g-6f+18-25=0$$\Rightarrow g^2+f^2+6g-6f=7 \qquad \cdots(i)$Since, point$(-1,-3)$lies on both circles, it satisfies both circle.∴ For circles$(-1)^2+(-3)^2+2g(-1)+2f(-3)+c=0$$\Rightarrow 1+9-2g-6f+c=0$$\Rightarrow -2g-6f+c=-10 \qquad \cdots(ii)$For circle $S'$,$(-1)^2+(-3)^2-6(-1)+6(-3)+2=0$$\Rightarrow 1+9+6-18+2=0$$\Rightarrow 0=0$Now, the slope of line joining $(-g,-f)$ and $(-1,-3)$ is equal to slope of line joining $(-1,-3)$ and $(3,-3)$$(-f+3) = \left(\frac{-3+3}{3+1}\right)(-g+1)$$\Rightarrow (-f+3)=0$$\Rightarrow f=3$Put $f=3$ into Eq. (i), we get$g^2+9+6g-6(3)$$\Rightarrow g^2+9+6g-18=7$$\Rightarrow g^2+6g-16=0$$\Rightarrow (g+8)(g-2)=0$$\Rightarrow g = -8 \text{ or } g=2 \text{ (not possible) }$Now, put $g=2, f=3$ into Eq. (ii), we get$-2(2)-6(3)+c = -10$$\Rightarrow c = -10+4+18=12$Now, $g+f+c=2+3+12=17$