Equation of family of circle passing through $S_1=0$ and point is $S_1+\lambda S_2=0$
$\begin{array}{r}\Rightarrow x^2+y^2+2 x+8 y-23+\lambda\left\{(x-1)^2+(y-2)^2\right\}=0 \\ \Rightarrow x^2+y^2+2 x+8 y-23+\lambda x^2+\lambda y^2-2 \lambda x-4 \lambda y+5 \lambda=0 \\ \Rightarrow \quad(\lambda+1) x^2+(\lambda+1) y^2+(2-2 \lambda) x \\ \quad+(8-4 \lambda) y+(5 \lambda-23)=0 \\ \Rightarrow \quad x^2+y^2+2 \frac{(1-\lambda)}{\lambda+1} x+2 \frac{(4-2 \lambda)}{\lambda+1} y+\frac{5 \lambda-23}{\lambda+1}=0\end{array}$
$\because$ radius $=\sqrt{10}$
$\begin{aligned} & \Rightarrow\left(\frac{1-\lambda}{\lambda+1}\right)^2+\left(\frac{4-2 \lambda}{\lambda+1}\right)^2-\left(\frac{5 \lambda-23}{\lambda+1}\right)=10 \\ & \Rightarrow \quad(1-\lambda)^2+(4-2 \lambda)^2-(5 \lambda-23)(\lambda+1)=10(\lambda+1)^2 \\ & \Rightarrow \quad 1+\lambda^2-2 \lambda+16+4 \lambda^2-16 \lambda-5 \lambda^2+18 \lambda+23 \\ & \quad=10 \lambda^2+10+20 \lambda \\ & \Rightarrow \quad 10 \lambda^2+20 \lambda-30=0 \Rightarrow \lambda^2+2 \lambda-3=0 \\ & \Rightarrow \quad(\lambda+3)(\lambda-1)=0 \Rightarrow \lambda=1,-3 \because \lambda \neq-1 \\ & \therefore \lambda=-3\end{aligned}$
$\therefore \quad$ Equation of circle is
$\begin{aligned} & x^2+y^2-4 x-10 y+19=0 \\ & a=-4, b=-10, c=19 \\ & \therefore \quad|a+b+c|=5 .\end{aligned}$