Any conic through the intersection has the form $S_1+\lambda S_2=0$:

$(x^2+2y^2-6x-12y+23)+\lambda (4x^2+2y^2-20x-12y+35)=0$.

For this to be a circle, the coefficients of $x^2$ and $y^2$ must be equal:

$1+4\lambda =2+2\lambda \Rightarrow \lambda =\frac{1}{2}$.

Substituting and dividing by 3:

$x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0$.

Wait, recompute: combining gives $3x^2+3y^2-16x-18y+(23+35/2)=0$, i.e. $x^2+y^2-\frac{16}{3}x-6y+\frac{81}{6}=0$.

Centre $(a,b)=\left(\frac{8}{3},3\right)$. Radius: $r^2=\frac{64}{9}+9-\frac{27}{2}=\frac{128+162-243}{18}=\frac{47}{18}$.

$ab+18r^2=\frac{8}{3}\cdot 3+47=8+47=55$.