A circle of radius $r$ passes through both foci of, and exactly four points on, the ellipse with equation $9x^2+25y^2=225$. The set of all possible values of $r$ is an interval $[a,b)$. What is $a+b$?

1. The ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ has $c=\sqrt{25-9}=4$, so the foci are $(\pm4,0)$. A circle through both foci has center $(0,k)$ and $r^2=k^2+16$, i.e. $x^2+y^2-2ky=16$.
2. The ellipse gives $x^2=25-\frac{25}9y^2$. Substituting: $25-\frac{16}9y^2-2ky=16$, i.e. $16y^2+18ky-81=0$, with roots $y=\dfrac{9(-k\pm r)}{16}$. They have opposite signs.
3. A point of the ellipse has $|y|\le3$; a root with $|y|<3$ gives two intersection points $(\pm x,y)$, and $|y|=3$ gives only one. For exactly four points we need both roots to have $|y|<3$.
4. For $k\ge0$ the positive root is smaller in size, and the negative root needs $\frac{9(r+k)}{16}<3$, i.e. $r<\frac{16}3-k$, i.e. $k^2+16<\frac{256}9-\frac{32}3k+k^2$, i.e. $k<\frac76$. By symmetry the condition is $|k|<\frac76$.
5. Hence $r=\sqrt{k^2+16}$ takes all values in $\left[4,\sqrt{\frac{49}{36}+16}\right)=\left[4,\frac{25}6\right)$, and $a+b=4+\frac{25}6=\boxed{\dfrac{49}6}$.