5. If $\alpha, \beta$ are the two real roots of the quadratic equation in $x$
$$
x^{2}-2 m x+1=m^{2}
$$

where $m$ is a real number, then the minimum value of $\alpha^{2}+\beta^{2}$ is ( ).
(A)3
(B) 4
(C) 1
(D) 2
5.C.

Since $\alpha+\beta=2 m, \alpha \beta=1-m^{2}$, therefore, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=6 m^{2}-2$.
Also, from $\Delta=4 m^{2}-4\left(1-m^{2}\right) \geqslant 0$, solving gives $m^{2} \geqslant \frac{1}{2}$. Thus, $\alpha^{2}+\beta^{2} \geqslant 6 \times \frac{1}{2}-2=1$.