B. Given that $a$, $b$, and $c$ are real constants, the quadratic equation in $x$
$$
a x^{2} + b x + c = 0
$$

has two non-zero real roots $x_{1}$ and $x_{2}$. Then, among the following quadratic equations in $x$, the one that has $\frac{1}{x_{1}^{2}}$ and $\frac{1}{x_{2}^{2}}$ as its two real roots is ( ).
(A) $c^{2} x^{2} + \left(b^{2} - 2 a c\right) x + a^{2} = 0$
(B) $c^{2} x^{2} - \left(b^{2} - 2 a c\right) x + a^{2} = 0$
(C) $c^{2} x^{2} + \left(b^{2} - 2 a c\right) x - a^{2} = 0$
(D) $c^{2} x^{2} - \left(b^{2} - 2 a c\right) x - a^{2} = 0$
B. B.

From the problem, we know that $a \neq 0$, and
$$
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} \neq 0 \text {. }
$$

Thus, $c \neq 0$,
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1}^{2} x_{2}^{2}}=\frac{b^{2}-2 a c}{c^{2}}, \\
\frac{1}{x_{1}^{2}} \cdot \frac{1}{x_{2}^{2}}=\frac{a^{2}}{c^{2}} .
\end{array}
$$

Therefore, by the relationship between roots and coefficients, the quadratic equation with $\frac{1}{x_{1}^{2}}, \frac{1}{x_{2}^{2}}$ as its two real roots is
$$
x^{2}-\frac{b^{2}-2 a c}{c^{2}} x+\frac{a^{2}}{c}=0 \text {, }
$$

which is $c^{2} x^{2}-\left(b^{2}-2 a c\right) x+a^{2}=0$.