4. Let $b>0, a^{2}-2 a b+c^{2}=0, b c>a^{2}$. Then the size relationship of the real numbers $a, b, c$ is ( ).
(A) $b>c>a$
(B) $c>a>b$
(C) $a>b>c$
(D) $b>a>c$
4. A.

From $bc > a^2$ and $b > 0$, we know $c > 0$.
From $2ab = a^2 + c^2$ and $b > 0$, we know $a > 0$.
From $a^2 - 2ab + c^2 = 0$, we know $b^2 - c^2 = (a - b)^2 \geqslant 0$.
Thus, $b \geqslant c$.
If $b = c$, from $a^2 - 2ab + c^2 = 0$, we know $a = b$.
Thus, $a = b = c$. This contradicts $bc > a^2$, so $b > c$.
From $b^2 > bc > a^2$, we know $b > a$;
Also, since $a^2 + c^2 = 2ab > 2a^2$, we know $c > a$.