Problem 171

Let M(x, y). From |OM|=2|MQ|, we get $$\sqrt {x^{2}+y^{2}}$$=2$$\sqrt {(x-2)^{2}+y^{2}}$$, which simplifies to x2+y2-$$\frac {16}{3}$$x+$$\frac {16}{3}$$=0. The center of this circle is ($$\frac {8}{3}$$,0) with a radius of $$\frac {4}{3}$$.

The problem is transformed into finding the intersection of the circle (x-a)2+y2=9 and the circle x2+y2-$$\frac {16}{3}$$x+$$\frac {16}{3}$$=0.

Hence, 3-$$\frac {4}{3}$$≤|a-$$\frac {8}{3}$$|≤3+$$\frac {4}{3}$$, which gives $$\frac {13}{3}$$≤a≤7. Since a>5, we have 5<a≤7.

Therefore, the answer is $\boxed{5<a\leq7}$.

This problem involves the relationship between a line and a circle and is of medium difficulty.