Inscribed circles under partial circles

Just to illustrate -

@Ronak Agarwal Great question. I think, though, that you may have to specify that the semicircle has radius $1$ to get the answer as posted. I initially solved assuming that the radius of the semicircle was some value $R,$ and in the end I obtained the general solution of $R.$

Really amazing problem ... Ronak !

And Brian Sir Why Radius should be 1 ?? What's wrong in my approach ... ?

If we Let $A(-R,0)$ & $B(R,0)$ & Center of given circle is $(0,0)$ , hence It is easy to see that , $C(R\cos { 2\theta } ,R\sin { 2\theta } )$ again let centre of other circle is $D(a,r)$ ... Now circles touches each .. hence

${ a }^{ 2 }+{ r }^{ 2 }=R+r\\ R={ a }^{ 2 }+{ r }^{ 2 }-r\quad \quad (1)$

Also drop Perpendicular from D to x-axis , name it as E , then in triangle ADE

$\cfrac { r }{ a } =\tan { \cfrac { \theta }{ 2 } } \\ a=r\cot { \cfrac { \theta }{ 2 } } \quad \quad (2)$

Equation of Tangent from A , to the new circle is $\displaystyle{L:\quad y-\tan { \theta } (x+R)=0\\ { \bot }^{ r }\quad from\quad D(a,r)\\ \left| \cfrac { a-\tan { \theta } (a+R) }{ \sqrt { 1+\tan ^{ 2 }{ \theta } } } \right| =r\\ (0,0)\quad gives\quad +ve\quad sign,hence\quad \\ remove\quad modulus\quad as\quad it\quad is:\\ R=\cfrac { a(1-\tan { \theta } )-r\sec { \theta } }{ \tan { \theta } } \quad \quad (3)\\ (1)\& (2)\& (3)\\ \\ \cfrac { \left( r\cot { \cfrac { \theta }{ 2 } } \right) \left( 1-\tan { \theta } \right) -r\sec { \theta } }{ \tan { \theta } } ={ r }^{ 2 }\cot ^{ 2 }{ \left( \cfrac { \theta }{ 2 } \right) } +{ r }^{ 2 }-r\\ \\ \cfrac { \left( \cot { \cfrac { \theta }{ 2 } } \right) \left( 1-\tan { \theta } \right) -\sec { \theta } }{ \tan { \theta } } ={ r }(1+\cot ^{ 2 }{ \left( \cfrac { \theta }{ 2 } \right) } )-1\\ r=f\left( \theta \right) =\left( \cfrac { \left( \cot { \cfrac { \theta }{ 2 } } \right) \left( 1-\tan { \theta } \right) -\sec { \theta } }{ \tan { \theta } } +1 \right) \sin ^{ 2 }{ \theta } }$

@Brian Charlesworth @Ronak Agarwal