Problem 129

From the given condition, we have $a\geqslant \frac {x^{2}+2xy}{x^{2}+y^{2}}$.

Consider $x^{2}+y^{2}=(1-m^{2})x^{2}+m^{2}x^{2}+y^{2}$ $(m > 0)$.

Applying the basic inequality, we have $(1-m^{2})x^{2}+m^{2}x^{2}+y^{2}\geqslant (1-m^{2})x^{2}+2mxy$, with equality holding if and only if $mx=y$.

Thus, $\frac {x^{2}+2xy}{x^{2}+y^{2}}\leqslant \frac {x^{2}+2xy}{(1-m^{2})x^{2}+2mxy}$.

When $1-m^{2}=m$, i.e., $m= \frac { \sqrt {5}-1}{2}$, $\frac {x^{2}+2xy}{x^{2}+y^{2}}$ attains its maximum value of $\frac {2}{ \sqrt {5}-1}= \frac { \sqrt {5}+1}{2}$.

Therefore, $a\geqslant \frac { \sqrt {5}+1}{2}$, so the minimum value of $a$ is $\boxed{\frac { \sqrt {5}+1}{2}}$.

This problem requires the application of the basic inequality to find the minimum value of $a$. Pay attention to the transformation and the conditions for equality to hold. This is a moderate-difficulty problem.