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I am unable to upload the image of my trials.

I assumed the radius of small circle is $x,$ horizontal distance between the centers of two circles is $y.$

I have joined the centers of the two circles and the length is $(5+x).$

I have drawn a vertical line from the center of the bigger circle to the center of the semi circle.

I have also drawn a horizontal line from the center of the small circle to the above line.

Then, by applying Pythagoras theorem, I get
$$(x+5)^2=(5-x)^2 + y^2.$$ I need one more equation to solve for $x.$

Intuitively I wonder if the radius of the small circle could be half of the big circle. I carefully constructed it and got $2.5 cm$ as the radius but I am not sure.

In addition to using Pythagorean theorem, one can use
circle inversion
to figure out the radius of the small circle (let's call it $r$).

Let $AB$ be the base of the semicircle. Let $O$ be its midpoint. Let $C$ be the contact
point between the small circle (blue) with the circular arc $AB$. Draw the line $OC$ and
let $D$ be its other intersection with the small circle. $CD$ will be a diameter of the small circle.

Perform a circle inversion with respect to the circle centered at $O$ with radius $10$.
The line $AB$ get mapped to itself. The big circle (green) get mapped to a line (green, dashed) parallel to $AB$ and at a distance $10$ from it. The small circle get mapped to a circle (blue, dashed) sandwiched between these two lines. So its diameter will be $10$. Let $D'$ be the image of $D$ under circle inversion. $CD'$ will be a diameter of the image of the small circle. We have

$$|CD'| = 10 implies |OD'| = |OC|+|CD'| = 10 + 10 = 2|OC|$$
Circle invert $OD'$ back to $OD$, we find
$$begin{align}|OD| = frac12|OC| &implies |CD| = |OC| - |OD| = frac12|OC|\
&implies r = frac12|CD| = frac14|OC| = frac52
end{align}
$$
The radius we seek is $frac52$. One half of that of the big circle and a quarter of that of the semicircle.

Let the radius of smaller circle be $displaystyle r$ and x-coordinate of its center$displaystyle ( C)$ is $displaystyle a$. As the circle is touching x-axis, so the ordinate of center of cicle is equal to radius of the circle,i.e., $displaystyle r$. Let the point of touching of semicircle and smaller circle be $displaystyle P_{1}$ $displaystyle ( x_{1}$$displaystyle ,y_{1})$. As the semicircle and smaller circle touch each other so $displaystyle P_{1}$, $displaystyle C,Origin$ are collinear.
begin{gather*}
therefore dfrac{r}{a} =dfrac{y_{1}}{x_{1}} ( 1)\
end{gather*}

And the distance between $displaystyle C$ and center of bigger circle is $displaystyle r+5$
begin{gather*}
( r-5)^{2} +a^{2} =( r+5)^{2}\
or 20r=a^{2} ( 2)
end{gather*}
Also the point $displaystyle P_{1}$ satisfies both the semicircle and the smaller circle.
begin{gather*}
therefore x^{2}_{1} +y^{2}_{1} =100 ( 3)\
And ( x_{1} -a)^{2} +( y_{1} -r)^{2} =r^{2}\
or x^{2}_{1} +y^{2}_{1} +a^{2} -2ax_{1} -2ry_{1} =0\
or 100+a^{2} -2ax_{1} -2ry_{1} =0 ( 4)
end{gather*}
Solving these four equations, we get $displaystyle r=2.5 units$

Let $R$ be the (known) radius of the large inscribed circle, $r$ the radius of the small inscribed circle, and $(x,r)$ the center of this small circle. Then one has the two equations
$$x^2+(R-r)^2=(R+r)^2,qquadsqrt{x^2+r^2}+ r=2R$$
in the two unknowns $r$ and $x$.

With hindsight - never any use, of course! - one can see that such a configuration must exist, because there exists an isosceles triangle whose sides are in the ratio of $3:2$ (whose height and angles one needn't know):