From rectangle to circle

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Both rectangles AEFG and ABNM are similar ( $\frac{AE}{AG} =\frac{AB}{AM}$ ) hence AF extended will meet the circle at N (A-F-N colinear).

Triangles AGF and AFM are similar right angled triangles $\angle AFG = \angle ANM = \angle FMA = \theta$

$\sin \theta = \frac{AG}{AF} = \frac{AF}{AM}$

Hence, $\frac{1}{\sqrt{5}} = \frac{10 \sqrt{5}}{R}$ and $\textbf {R = 50}$

Let R = radius of the circle. Then there will be eight points like F on the circle. Take the one shown with coordinates F = (-R+20, R-10) with center O as the origin.

$(-R + 20)^2 + (R - 10)^2 = R^2$

$R^2 -60R + 500 = 0$ giving R = 10 or 50. R= 10 too gives an interesting solution! In that case, the rectangle AEFG overlaps the circle and F is the midpoint of the right edge of the square.

But the one we need is R = 50 giving the area to be $50^2 \pi = 7853.98$

$\color{#3D99F6}{AG=EF=10}, \color{#20A900}{AG=AE=20}$

Let $r$ be the radius of the circle.

$\begin{aligned} OF^{2}= r^{2} & =(r-\color{#3D99F6}{EF})^{2}+(r-\color{#20A900}{GF})^{2} \\ r^{2} & =(r-\color{#3D99F6}{AG})^2+(r-\color{#20A900}{AE})^{2} \\ r^{2} & =(r-\color{#3D99F6}{10})^2+(r-\color{#20A900}{20})^{2} \\ r^{2} & =r^{2}-20r+100+r^{2}-40r+400 \\ 0 & =2r^{2}-60r+500-r^{2} \\ 0 & =r^{2}-60r+500 \\ 0 & =(r-10)(r-50) \\ & r=10^\color{#D61F06}{*} \text{ or } r=50 \quad \quad \small \color{#D61F06}{\text{* not available because smaller than AE}} \end{aligned}$

$Area= \pi r^{2}=\pi 50^{2}=\boxed{7853.9816}$

Let 20=rcosx and 10=r(1-sinx). Now we have cosx=2(1-sinx). Simplifying yields 2sinx+cosx=2. Let cosx=sqrt(1-x^2), where x=sinx, a quadratic can be solved, yielding sinx=4/5. Now 10=r/5 and r=50, hence area is 2500pi.