Circle/ Ellipse geometry problem from Ramanujan Does someone have detail of the problem and his solution? I saw it in youtube. A rough hand sketch remade seeing YouTube video SR geometry problem  (17 min 51 sec). I have no access to the  TIFR book cited.

Thanks in advance.
The original image is here :

 A: Background
This can be found in Mathematics Magazine, Vol. 51, No. 3 (May, 1978), pg 161:



A similar sketch can be found in the Manuscript Book 1 of Srinivasa Ramanujan,  pg. 54:



The Problem
This can be found in the Journal of the Indian Mathematical Society, V, 1913, pg 24:

Let $PQR$ be a circle with center $O$, of which a diameter is $PR$.
Bisect $PO$ at $H$ and let $T$ be the point of trisection of $OR$
nearer $R$. Draw $TQ$ perpendicular to $PR$ and place the chord $RS =TQ$.
Join $PS$, and draw $OM$ and $TN$ parallel to $RS$. Place a chord $PK= PM$, and draw the tangent $PL = MN$. Join $RL, RK$ and $KL$. Cut off $RC = RH$. Draw $CD$ parallel to $KL$, meeting $RL$ at $D.$

Then the square on $RD$ will be equal to the circle $PQR$
approximately. For
$$RS^2=\frac{5}{36}d^2,$$
where $d$ is the diameter of the circle. Therefore
$$PS^2=\frac{31}{36}d^2.$$
But $PL$ and $PK$ are equal to $MN$ and $PN$ respectively. Therefore
$$PK^2=\frac{31}{144}d^2,\quad \text{and} \quad PL^2=\frac{31}{324}d^2.$$
Squaring the circle
Hence
$$RK^2=PR^2-PK^2=\frac{113}{114}d^2,$$
and
$$RL^2=PR^2+PL^2=\frac{355}{324}d^2.$$
But
$$\frac{RK}{RL}=\frac{RC}{RD}=\frac{3}{2}\sqrt{\frac{113}{355}},$$
and
$$RC=\frac{3}{4}d.$$
Therefore
$$RD=\frac{d}{2}\sqrt{\frac{355}{113}}=r\sqrt{\pi},\quad \text{very nearly}$$

Perhaps this will help you in your studies
