A geometry problem by Ajay Sambhriya

Relevant wiki: Pythagorean Theorem

Flip the figure horizontally and let $A$ be the origin, $AC$ along the $x$ -axis and $AB$ along the $y$ -axis. Let $D$ be the center of the grey circle. We note that $DE= r$ , the radius of the grey circle and that

$\begin{aligned} AG & = 7 \\ {\color{#3D99F6}AD} + DG & = 7 & \small \color{#3D99F6} \text{By Pythagorean theorem} \\ {\color{#3D99F6}\sqrt{AE^2+DE^2}} + r & = 7 \\ \sqrt{d^2+r^2} + r & = 7 \\ \sqrt{d^2+r^2} & = 7 - r & \small \color{#3D99F6} \text{Squaring both sides} \\ d^2+r^2 & = 49-14r+r^2 \\ \implies d^2 & = 49-14r \end{aligned}$

Note that the equation for the grey circle is:

$\begin{aligned} (x-d)^2 + (y-r)^2 & = r^2 & \small \color{#3D99F6} \text{For point }F(0,1) \\ {\color{#3D99F6}d^2} + (1-r)^2 & = r^2 & \small \color{#3D99F6} \text{Note that } d^2 = 49-14r \\ 49-14r + 1 - 2r + r^2 & = r^2 \\ 50-16r & = 0 \\ \implies r & = \boxed{\dfrac {25}8} \end{aligned}$

Let $|AE| = d$ , and let $|DE| = r$ be the radius of the grey circle. With $A$ being the origin and $AC$ lying on the $x$ -axis, the equation of this circle is then

$(x + d)^{2} + (y - r)^{2} = r^{2}$ . Now as point $F$ lies on this circle and $|AF| = 1$ we have that $(0,1)$ satisfies this equation, i.e., that

$(0 + d)^{2} + (1 - r)^{2} = r^{2} \Longrightarrow d^{2} + 1 - 2r + r^{2} = r^{2} \Longrightarrow d = \sqrt{2r - 1}$ .

Next, we note that $AD$ extended will pass through the upper point $P$ of intersection between the grey circle and blue quarter-circle. Then as $|DP| = r$ and $|AP| = |AB| = 7$ we have that $|AD| = |AP| - |DP| = 7 - r$ . Looking then at the right triangle $\Delta AED$ , by the Pythagorean Theorem we have that

$|AD|^{2} = |AE|^{2} + |DE|^{2} \Longrightarrow (7 - r)^{2} = d^{2} + r^{2} = 2r - 1 + r^{2} \Longrightarrow 49 - 14r + r^{2} = 2r - 1 + r^{2} \Longrightarrow 16r = 50 \Longrightarrow \boxed{r = \dfrac{25}{8}}$ .