You take one side, I'll take the other

Orient the two circles on an $xy$ -grid centered at the origin. Let the rectangle be symmetric about the $x$ -axis, with the vertices on the radius $4$ circle to the right of those on the radius $3$ circle.

Now let $y$ be the vertical dimension of the rectangle and $x$ the horizontal dimension. Then by Pythagoras we have that

$x = \sqrt{3^{2} - (\frac{y}{2})^{2}} + \sqrt{4^{2} - (\frac{y}{2})^{2}}$ .

(Note that it should be clear that the left side of the rectangle of largest area will be to the left of the $y$ -axis. If it's not clear, then for now just assume this to be the case.)

The area $A$ of the rectangle is then

$A = xy = \dfrac{1}{2}*y*[\sqrt{36 - y^{2}} + \sqrt{64 - y^{2}}]$ .

We then need to find where $\dfrac{dA}{dy} = 0$ to find where $A$ is maximized. Doing this we find that

$\dfrac{dA}{dy} = (\dfrac{1}{2})[\sqrt{36 - y^{2}} + \sqrt{64 - y^{2}} - \dfrac{y^{2}}{\sqrt{36 - y^{2}}} - \dfrac{y^{2}}{\sqrt{64 - y^{2}}}] = 0$ when

$y^{2} = \sqrt{36 - y^{2}} \sqrt{64 - y^{2}} \Longrightarrow y^{4} = 2304 - 100y^{2} + y^{4} \Longrightarrow y^{2} = \dfrac{2304}{100} \Longrightarrow y = \dfrac{24}{5}$ .

Plugging this back into the equation for $x$ yields $x = 5$ , and so $x - y = \dfrac{1}{5}$ . Thus $a = 1, b = 5$ and $a + b = \boxed{6}$ .