The general equation of a circle is $x^2+y^2+2Gx+2Fy+C=0$. For $S_1$, $G_1=-2, F_1=-3, C_1=5$. For $S_2$, $G_2=g, F_2=f, C_2=c$. The radical axis of $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. $(x^2+y^2-4x-6y+5) - (x^2+y^2+2gx+2fy+c) = 0$ $(-4-2g)x + (-6-2f)y + (5-c) = 0$. This equation represents the line $x+y-1=0$. For these two equations to represent the same line, their coefficients must be proportional. Let the proportionality constant be $k$: $-4-2g = k \cdot 1 \quad \ldots (1)$ $-6-2f = k \cdot 1 \quad \ldots (2)$ $5-c = k \cdot (-1) \quad \ldots (3)$ From (1) and (2): $-4-2g = -6-2f \Rightarrow 2f-2g = -2 \Rightarrow f-g=-1 \quad \ldots (A)$ From (1) and (3): $-4-2g = -(5-c) = c-5 \Rightarrow c = 1-2g \quad \ldots (B)$ Circle $S_2$ passes through the point $(1,1)$. Substituting $x=1, y=1$ into the equation of $S_2$: $1^2+1^2+2g(1)+2f(1)+c=0 \Rightarrow 2+2g+2f+c=0 \quad \ldots (C)$ Now, substitute equations (A) and (B) into (C): $2+2g+2(g-1)+(1-2g)=0$ $2+2g+2g-2+1-2g=0$ $2g+1=0 \Rightarrow g = -1/2$. Using $g=-1/2$ in (A): $f = g-1 = -1/2-1 = -3/2$. Using $g=-1/2$ in (B): $c = 1-2(-1/2) = 1+1 = 2$. Finally, we need to find $g+f+c = (-1/2) + (-3/2) + 2 = -4/2 + 2 = -2+2 = 0$.