4. If the equations $x^{2}+2 a x+b^{2}=0$ and $x^{2}+2 c x-$ $b^{2}=0$ have a common root, and $a, b, c$ are exactly the three sides of a triangle, then this triangle is ( ).
(A) Acute triangle
(B) Obtuse triangle
(C) Right triangle with $a$ as the hypotenuse
(D) Right triangle with $c$ as the hypotenuse
4. (C).
$x^{2}+2 a x+b^{2}=0, x^{2}+2 c x-b^{2}=0$ subtract to get $2 a x-2 c x+2 b^{2}=0$.
Obviously, $c \neq a$. (Otherwise, if $a=c$, then we have $b=0$, contradicting the assumption $b$
$>($ transition)
Thus $x=\frac{b^{2}}{c-a}$. This is the common root of the two equations.
Substitute $x=\frac{b^{2}}{c-a}$ into one of the equations, for example, into $x^{2}+$ $2 c x=b^{2}$, we get
$$
\begin{array}{l}
\left(\frac{b^{2}}{c-a}\right)^{2}+2 c \cdot \frac{b^{2}}{c-a}=b^{2}, \\
b^{4}+2 c(c-a) b^{2}=b^{2}(c-a)^{2},
\end{array}
$$

which simplifies to $b^{2}+2 c(c-a)=(c-a)^{2}$.
Simplifying further, we get $b^{2}+c^{2}=a^{2}$.
This triangle is a right triangle with $a$ as the hypotenuse.