This approach uses the regular decagon.

Assuming, as indicated in OP's second figure, that the left and right small circles are tangent to the extended diameter of the larger circle at $E$ and $F$, then $CB$ is the side of a regular decagon, $\triangle ABC$ is $36^o-72^o-72^o$, and$$\frac{AB}{BC}=\phi=\frac{1+\sqrt 5}{2}$$

Letting radius $BD=1$, then$$\frac{AB}{BD}=\frac{1+\sqrt 5}{1}$$and$$\frac{AD}{BD}=\frac{1+\sqrt 5-1}{1}=\frac{\sqrt 5}{1}$$Thus the larger circle has five times the area of the smaller circle.

Following the suggestion of @Blue, this approach uses the regular decagon.

Assuming, as indicated in OP's second figure, that the left and right small circles are tangent to the extended diameter of the larger circle at $E$ and $F$, then $CB$ is the side of a regular decagon, $\triangle ABC$ is $36^o-72^o-72^o$, and$$\frac{AB}{BC}=\phi=\frac{1+\sqrt 5}{2}$$

Letting radius $BD=1$, then$$\frac{AB}{BD}=\frac{1+\sqrt 5}{1}$$and$$\frac{AD}{BD}=\frac{1+\sqrt 5-1}{1}=\frac{\sqrt 5}{1}$$Thus the larger circle has five times the area of the smaller circle.