Two circles lie outside the regular octagon $ABCDEFGH$. The first is tangent to $\overline{AB}$, and the second is tangent to $\overline{EF}$. Both are tangent to lines $BC$ and $HA$. What is the ratio of the radius of the second circle to the radius of the first circle?

1. Let the side of the octagon be $2$. Each exterior angle of a regular octagon is $45^\circ$, so lines $HA$ and $BC$ meet at a point $P$ below $AB$ with $\angle APB=90^\circ$. Triangle $PAB$ is an isosceles right triangle with hypotenuse $AB=2$, so $P$ is at distance $1$ from $AB$.
2. Both circles lie inside the $90^\circ$ angle at $P$, so their centers are on the bisector of that angle (the perpendicular bisector of $AB$), and a circle whose center is at distance $t$ from $P$ has radius $\dfrac t{\sqrt2}$.
3. The first circle is the incircle of $\triangle PAB$: its center is at distance $\sqrt2\,r_1$ from $P$ and it touches $AB$ at distance $1$ from $P$, so $\sqrt2\,r_1+r_1=1$, giving $r_1=\sqrt2-1$.
4. The distance between the opposite sides $AB$ and $EF$ of the octagon is $2(1+\sqrt2)$, so $EF$ is at distance $1+2(1+\sqrt2)=3+2\sqrt2$ from $P$. The second circle lies beyond $EF$, so $\sqrt2\,r_2=(3+2\sqrt2)+r_2$, giving $r_2=\dfrac{3+2\sqrt2}{\sqrt2-1}=7+5\sqrt2$.
5. The ratio is $\dfrac{r_2}{r_1}=\dfrac{7+5\sqrt2}{\sqrt2-1}=(7+5\sqrt2)(\sqrt2+1)=\boxed{17+12\sqrt2}$, which equals $(1+\sqrt2)^4\approx33.97$.