$17 \cdot 182$ If the equations $x^{2}+2 a x+b^{2}=0$ and $x^{2}+2 c x-b^{2}=0$ have one common root, and $a, b, c$ are the three sides of a triangle, then this triangle is
(A) an acute triangle.
(B) an obtuse triangle.
(C) a right triangle with $c$ as the hypotenuse.
(D) a right triangle with $a$ as the hypotenuse.
(3rd "Jinyun Cup" Junior High School Mathematics Invitational, 1986)
［Solution］Let $\alpha$ be the common root of the two equations. According to the problem, we have

i.e. $\square$
$$
\begin{array}{c}
\alpha^{2}+2 a \alpha+b^{2}=\alpha^{2}+2 c \alpha-b^{2}, \\
b^{2}=(c-a) \alpha .
\end{array}
$$

Given that $b \neq 0$, hence $c-a \neq 0$.
Then $\alpha=\frac{b^{2}}{c-a}$. Substituting into the equation, we get
$$
\frac{b^{4}}{(c-a)^{2}}+\frac{2 a b^{2}}{c-a}+b^{2}=0,
$$

i.e. $\square$ $b^{2}+c^{2}-a^{2}=0$ or $a^{2}=b^{2}+c^{2}$. Therefore, the answer is $(D)$.