The equation of a family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. So, the equation of the required circle is $(x^2+y^2-8x-2y+7) + \lambda(x^2+y^2-4x+2y-1) = 0$. This circle passes through the point $(1,2)$. Substituting $x=1$ and $y=2$ into the equation: $(1^2+2^2-8(1)-2(2)+7) + \lambda(1^2+2^2-4(1)+2(2)-1) = 0$ $(1+4-8-4+7) + \lambda(1+4-4+4-1) = 0$ $(12-12) + \lambda(9-5) = 0$ $0 + 4\lambda = 0 \Rightarrow \lambda = 0$. Substituting $\lambda=0$ back into the family equation, we get $x^2+y^2-8x-2y+7=0$. This is the equation of the first circle, which means the point $(1,2)$ actually lies on the first circle itself.