Radius of semicircle inscribing $5$ congruent circles of radius $r$ There are $5$ congruent circles with radius $r$ inscribed in a semicircle as shown in the below diagram. What is the radius of the semicircle?

My attempt:
If the min distance between circle $C_2$ and $C_3$ is $2x$, $C_2C_3 = 2 (r + x)$
$C_2C_5 = C_3C_5 = 2r$
Perpendicular from $C_5$ to the line $C_2C_3$ = $r+h$ then
$(r+h)^2 + (r+x)^2 = (2r)^2$
If I get one more equation in $x, h, r$, I can find $x$ and $h$ in terms of $r$.
If we extend $C_1C_4$ and $C_3C_5$, they will meet at point $M$, with $MC_2$ perpendicular to line $C_1C_2$. While I know $C_1C_2 = 2(r+x)$, I am not sure I can express $C_2M$ in terms of $r$ and $h$.
I also tried using trigonometry but could not find the radius of the bigger circle.
I would appreciate any directional help or solution.
 A: 
Introduce $R$ as the radius of the semicircle.
From triangle $OC_2C_4$:
$$OC_4^2=OC_2^2+C_2C_4^2-2OC_2OC_4\cos(90^\circ + \alpha)$$
$$(R-r)^2=r^2+(2r)^2+2r(2r)\sin\alpha$$
$$R^2-2Rr=4r^2+4r^2\sin\alpha\tag{1}$$
From triangle $C_1C_2C_4$:
$$C_1C_2=4r\cos\alpha$$
From triangle $OC_1C_2$:
$$OC_1^2=OC_2^2+C_1C_2^2$$
$$(R-r)^2=r^2+(4r\cos\alpha)^2$$
$$R^2-2Rr=16r^2\cos^2\alpha\tag{2}$$
From (1) and (2):
$$4r^2(1+\sin\alpha)=16r^2\cos^2\alpha$$
$$1+\sin\alpha=4(1-\sin^2\alpha)$$
$$4\sin^2\alpha+\sin\alpha-3=0$$
Solve this quadratic equation and take only positive solution:
$$\sin\alpha=\frac 34\tag{3}$$
Now replace (3) into (1):
$$R^2-2Rr=4r^2(1+\frac 34)$$
$$R^2-2Rr-7r^2=0$$
This quadratic equation has only one positive solution:
$$R=r(1+2\sqrt 2)$$
A: 
Once you prove $OM = ON$ which is easy to establish (see my alternate solution), the method you are trying can be completed using $\triangle OJM, \triangle ONI, \triangle CMN$.
WLOG, assume $r = 1$
$OM^2 = 4(1+x)^2+1$
$ON^2 = (1+x)^2+(2+h)^2$
As $OM = ON$,
$3(1+x)^2 + 1 = (2+h)^2$ ...(i)
$CN^2 = (1+x)^2+(1+h)^2$
i.e. $(1+x)^2 = 4 - (1+h)^2$ ...(ii)
Substituting (ii) in (i) and solving, $OM = 2\sqrt2$.
So ratio of radius of semicircle to radius of smaller circle $= 2\sqrt2+1$.
Alternate solution using similar triangles -
say, $\angle MOJ = \theta$
If midpoint of $MN$ is point $P$, $\angle MOP = \theta, \angle MON = 2\theta$
$\angle MNO = \angle NMO = 90^0 - \theta, \angle MOQ = 90^0-\theta$
So, $\triangle OMN \sim \triangle QOM$
$\dfrac {OM}{MN} = \dfrac {QM}{OM}$
$OM = \sqrt{(QM)(MN)} = 2\sqrt2 r$
Desired radius $= 2\sqrt2 r + r = r(2\sqrt2+1)$
