(4) Given the equation in $x$, $x^{2}-2 a x+a^{2}-4 a=0$, has at least one complex root with a modulus of 3, then the possible values of the real number $a$ are ( ).
(A) 1,9
(B) $-1,9,2-\sqrt{13}$
(C) $1,9,2+\sqrt{13}$
(D) $1,9,2-\sqrt{13}$
(4) Write the equation as $(x-a)^{2}=4 a$. When $a \geqslant 0$, the equation has real roots because the modulus of the real root is 3, so one root of the equation is 3 or -3. Substituting, from $(a \pm 3)^{2}=4 a$, we get $a=1$ or 9; when $a<0$, we get $x=a \pm 2 \sqrt{|a|} \mathrm{i}$, so $|x|^{2}=a^{2}-4 a=9$, we get $a=2-\sqrt{13}$. Therefore, the answer is D.