$13 \cdot 25$ Let $y=(x-a)^{2}+(x-b)^{2}$, where $a, b$ are constants. When $y$ has a minimum value, then $x$ equals
(A) $\frac{a+b}{2}$.
(B) $a+b$.
(C) $\sqrt{a b}$.
(D) $\sqrt{\frac{a^{2}+b^{2}}{2}}$.
(E) $\frac{a+b}{2 a b}$.
(15th American High School Mathematics Examination, 1964)
[Solution] According to the problem, we have $y=2 x^{2}-2(a+b) x+a^{2}+b^{2}$. Therefore, when $x=\frac{a+b}{2}$, the function $y$ has a minimum value. Hence, the correct choice is $(A)$.