jmc

We have $a+8c=4+7b$ and $8a-c=7-4b$. Squaring both equations and adding the results yields $$(a+8c{)}^2+(8a-c{)}^2=(4+7b{)}^2+(7-4b{)}^2.$$Expanding gives $65(a^2+c^2)=65(1+b^2)$. So $a^2+c^2=1+b^2$, and $a^2-b^2+c^2=\boxed{1}$.