$1 \cdot 37$ If $x y=b$ and $\frac{1}{x^{2}}+\frac{1}{y^{2}}=a$, then $(x+y)^{2}$ is equal to
(A) $(a+2 b)^{2}$.
(B) $a^{2}+b^{2}$.
(C) $b(a b+2)$.
(D) $a b(b+2)$.
(E) $\frac{1}{a}+2 b$.
(9th American High School Mathematics Examination, 1958)
[Solution] $(x+y)^{2}=x^{2}+y^{2}+2xy=x^{2}y^{2}\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{xy}\right)$ $=b^{2} \cdot\left(a+\frac{2}{b}\right)=ab^{2}+2b$.

Therefore, the answer is $(C)$.