Those two semicircles

$CO=OF=AO=6, DG=GE \Rightarrow GE=3.5, OG=OF-GE-EF=6-3.5-2=0.5$

$GE^2 + h^2 = r^2$

$OO_2^2+AO_2^2=AO^2 \Rightarrow h^2 + OG^2+r^2=AO^2$

$3.5^2 + h^2 = r^2$

$h^2 + r^2 + 0.5^2=6^2$

Solving these two equations we get value of $r=2\sqrt{6} \Rightarrow AB=4\sqrt{6} \approx \boxed{9.7979}$

Let the center of the blue semicircle be the origin $O(0,0)$ , the center of the red semicircle be $C$ and its radius be $r$ .

Since the triangle with vertices at $C$ , $(-3,0)$ and $(4,0)$ is an isosceles triangle. The $x$ coordinate of $C$ is 0.5. Let the $y$ coordinate of $C$ be $a.$

Then the equation of the blue semicircle is $x^2 + y^2 = 36 \ ...(1)$ and that of the red semicircle:

$\begin{aligned} (x-0.5)^2 + (y-a)^2 & =\color{#3D99F6} r^2 & \small \color{#3D99F6} \text{Note that }r^2 = a^2 + 3.5^2 \\ x^2 - x + 0.25 + y^2 - 2ay + a^2 & = \color{#3D99F6} a^2 + 12.25 \\ x^2 - x + y^2 - 2ay & = 12 & ...(2) \end{aligned}$

The two points of intersection $A$ and $B$ satisfy $(1)-(2): \ x + 2ay = 24$ , which is a straight line joining $A$ and $B$ . For line $AB$ to be the diameter of the red semicircle, it must pass through $C(0.5, a)$ . Therefore,

$\begin{aligned} 0.5 + 2a^2 & = 24 \\ \implies a & = \sqrt{\frac {24-0.5}2} = \frac {\sqrt {47}}2 \\ \implies r^2 & = \frac {47}4 + \left(\frac 72\right)^2 = 24\end{aligned}$

Therefore, the length of $AB$ , the diameter of the red semicircle is $2r = 2\sqrt {24} \approx \boxed{9.7980}$ .

Let $O$ be the center of the red circle, $P$ be the center of the blue circle, $DG$ be the given diameter of the blue circle, and $E$ and $F$ be the intersection points of the two semicircles, as shown below. Extend $AB$ and $DG$ so that they intersect at $C$ , and draw $PB$ and $OE$ .

The diameter of the blue circle $DG = 3 + 7 + 2 = 12$ , so the radius is $PD = PB = 6$ .

By the intersecting secant theorem on the blue circle, $AB \cdot AC = CD \cdot CG$ or $AB(AB + BC) = CD(CD + 12)$ . By the intersecting secant theorem on the red circle, $AB \cdot AC = CE \cdot CF$ or $AB(AB + BC) = (CD + 2)(CD + 9)$ . Then $AB(AB + BC) = CD(CD + 12) = (CD + 2)(CD + 9)$ , which means $CD = 18$ , and $AB(AB + BC) = 540$ .

Using the law of cosines on $\triangle BCP$ , $\cos C = \frac{CP^2 + BC^2 - BP^2}{2 \cdot CP \cdot BC}$ , or $\cos C = \frac{24^2 + BC^2 - 6^2}{2 \cdot 24 \cdot BC}$ , or $\cos C = \frac{BC^2 + 540}{48 \cdot BC}$ . Using the law of cosines on $\triangle CEO$ , $\cos C = \frac{CE^2 + CO^2 - EO^2}{2 \cdot CE \cdot CO}$ , or $\cos C = \frac{20^2 + (\frac{1}{2}AB)^2 - (\frac{1}{2}AB + BC)^2}{2 \cdot 20 \cdot \frac{1}{2}AB}$ , or since $AB(AB + BC) = 540$ , $\cos C = \frac{47 \cdot BC}{BC^2 + 540}$ . Then $\cos C = \frac{BC^2 + 540}{48 \cdot BC} = \frac{47 \cdot BC}{BC^2 + 540}$ , which means $BC = 2\sqrt{141} - 2\sqrt{6}$ , and since $AB(AB + BC) = 540$ , $AB = 4\sqrt{6} \approx \boxed{9.7979}$ .