2002 AMC 10B Problem 20

Rewrite the equations as $a + 8c = 4 + 7b$ and $8a - c = 7 - 4b.$ Squaring both and adding, $$\begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2. \end{aligned}$$ The left side expands to $65a^2 + 65c^2$ (the $ac$ terms cancel), and the right side expands to $65 + 65b^2$ (the $b$ terms cancel). So $$65(a^2 + c^2) = 65(1 + b^2),$$ giving $a^2 - b^2 + c^2 = 1.$ Thus, the correct answer is B.