Find the radius of three equal circles, with radius $r$, inscribed in a rectangle. How can we find the radius of the small circles with the same given radius $r$. Three equal circles, with radius $r$, are inscribed in a rectangle in a way only one of them touches the others two, as the figure indicates. The circles centres form an isosceles triangle, is there any theorem can help in finding the value of $r$? Thanks for your help.

 A: Assume it is a big circle with radius $R$ that the smaller circles are tangent to. As fleablood commented, if that white arc is not circular (while maintaining mirror symmetry, as per the isosceles requirement) then the answer would be different.
Note that the given lengths of the "frame" forms the $3$-$4$-$5$ Pythagorean triple. 
In the diagram below, the center of the big circle, point $O$, can be found by the perpendicular bisector of $\overline{PC}$ which length is $|PC|=10$. 
With the similarity $\triangle PBC \sim \triangle OPQ~$ we have
$$R = |PO| = |PQ| \cdot\frac{|PC|}{|BC|} = |PQ| \cdot\frac53 = \frac{25}3$$

Next, please consult the diagram below to obtain the desired small circle radius $r$.
Observe that the height is related to the small radius:
$$ |BC| = |PF|+|FJ|+|EH| = 2r + |FJ|~,$$
then for $|FJ|$, from the similarity $\triangle OFG \sim \triangle EFJ~$ we have
$$|FJ| = |FE| \cdot \frac{|FJ|}{|FE|} = |FE| \cdot \frac{|FG|}{|FO|} = 2r \cdot \frac{r}{R-r} \\
\implies |BC| = 2r \Bigl( 1+\frac{r}{R-r} \Bigr) \qquad \text{, with}~~|BC| = 6$$
Solve for $r$ with $R = \dfrac{25}3$ obtained above, we arrive at $r = \dfrac{75}{34} \approx 2.206$.

Note that point $E$ being the center of the small circle on the right, is different from midpoint of $\overline{PC}$ point $Q$ in the previous diagram (not shown here).
$$\begin{aligned}
2r \Bigl( 1+\frac{r}{R-r} \Bigr) = 6 \quad &\implies r+\frac{r^2}{R-r}  = 3\\
&\implies rR -r^2 + r^2  = 3R-3r \\
&\implies r(R+3)  = 3R \\
&\implies r= \frac{3R}{3+R} = \frac{3 \cdot 25/3}{3+25/3}  = \frac{75}{34}
\end{aligned}$$
A: 
\begin{align}
R^2&=|AH|^2-|HO|^2=\tfrac{a^2}4+(R-b)^2
,\\
R&=\frac{a^2+4\,b^2}{8\,b}
=
\frac{25}3
,\\
|HE|&=|HK|+|KO_0|+|O_0E|
,\\
b&=r_0+\sqrt{4\,r_0^2-((R-r_0)^2-(R-b+r_0)^2 )}+r_0
,\\
r_0&=\frac{R\,b}{2\,R+b}
=\frac b2\cdot\frac{(a^2+4\,b^2)}{a^2+8\,b^2}
=\frac{75}{34}
\approx 2.20588
.
\end{align} 
A: If the white curve is a circle the you have three points of the big circle (assume symmetry: Big circle has $(-8,0),(0,6),(8,0)$) so the equation (and radius) can be figured. 
Big Circle is center at $(0, -A)$ with radius $R$ and $8^2 + A^2 = R^2$ and $0^2 + (A+8)^2 = R^2$.
The three little circles have equations $(a_i - x)^2 + (b_i-y)^2 = r^2$.  $a_1=-a_3;b_1=b_3; a_2 = 0$. circle 1: has the three points $(-w,0), (-v,z);$ (which satisfies $v^2 + (z+A)^2 = R^2$) and $(-m,k)$.  
Circle 2: has three point $(-m,k),(m,k)$ and $(0,6)$.  The point $(-m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
And circle 3 has three points $(w,0),(v,z),(m,k)$.The point $(m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
So solve the following equations.
Big Circle:
$(8-0)^2 + (0+A)^2 = R^2$
$(0-0)^2 + (6 + A)^2 = R^2$
$(m-0)^2 + (k+A)^2 = R^2$
$(v-0)^2 + (z+A)62 = R^2$
Circle 2:
$(m-0)^2 + (k-b_2)^2 = (0-0)^2 + (6-b_2)^2 = r^2$.
Circle 3:
$(a_3-w)^2 + (b_3-0)^2 = (a_3 - v)^2 + (b_3 - z)^2=(a_3-m)^2 + (b_3-k)^2 = r^2$
Colinear tangent of circle 2 and circle 3
$m = \frac {a_3}2$ and $k = \frac {b_2 + b_3}2$.
