Golden Ratio in Three Regular Pentagons II

The problem and all variants can be solved by explicit coordinates of the involved points, using $\cos 36^{\circ}= \displaystyle \frac{\varphi}{2}.\,$ For example, when $ABCDE\,$ is the unit pentagon with $A=(1,0),\,$ the circle of the problem is centered at $\displaystyle \left(\frac{3-\sqrt{5}}{2}, 0\right)\,$ with squared radius $4-\sqrt{5}\,$ and $\displaystyle T = \left(3 - 3\frac{\varphi}{2}, \sqrt{3-\varphi}\left(\varphi-\frac{1}{2}\right)\right),\,$ $\displaystyle B=\left(\frac{\varphi-1}{2}, \frac{\varphi}{2}\cdot\sqrt{3-\varphi}\right),\,$ $F=(1, \varphi\cdot\sqrt{3-\varphi}).$

The configuration sports several occurences of the Golden Ratio. The applet below highlights seven of them. #5 is the original one. ##1-3 are rather traditional and simple, #4 is intermediate, ##5-7 are difficult. (I apologize for a change of notations.)

The problem has been kindly posted on the CutTheKnotMath facebook page by Tran Quang Hung.