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One circle is inscribed inside a unit square such that all sides of square are tangent to it.
A second circle is at the upper-left corner of the square such that it is tangent to the two sides of the square and touches the first circle externally.
A third circle touches both circles externally and the upper side of the square is tangent to the third circle. (it is on the right side of the second circle.)
Find the radius of the 3rd circle.
Let the three radii be $r_1$, $r_2$ and $r_3$. From matching various lengths, the following relationships among the radii hold,
$$\sqrt{2}r_1=r_1+(1+\sqrt{2})r_2\tag{1}$$
$$(r_1+r_3)^2-(r_1-r_3)^2=a^2\tag{2}$$ $$(r_2+r_3)^2-(r_2-r_3)^2=b^2\tag{3}$$ $$r_1-r_2=a+b\tag{4}$$
Solve (1) for the radius of the second circle, $$r_2=(3-2\sqrt{2})r_1$$
Simplify (2) and (3) to get,
$$r_3=\frac{a^2}{4r_1}\tag{5}$$ $$\frac{a^2}{b^2}=\frac{r_1}{r_2}\tag{6}$$
Combine (6) and (4) to get $$a=\sqrt{r_1}(\sqrt{r_1}-\sqrt{r_2})\tag{7}$$
Then, plug (7) into (5) to obtained the radius of the third circle,
$$r_3=\frac 14 (\sqrt{r_1}-\sqrt{r_2})^2 $$
Use the result of the second radius $r_2$ derived above to express $r_3$ in terms of $r_1 = \frac 12$,
$$r_3=\frac 12 (3-2\sqrt{2})r_1=\frac 14 (3-2\sqrt{2})$$
As @Jaap Scherphuis commented, this is a special case of Descartes' Theorem
\begin{align}
k_3&=k_1+k_2+k_4+2\sqrt{k_1k_2+k_2k_4+k_4k_1}
=k_1+k_2+2\sqrt{k_1k_2}
,
\end{align}
where
\begin{align} k_1&=1/r_1 ,\\ k_2&=1/r_2 ,\\ k_3&=1/r_3 ,\\ k_4&=0 , \end{align}
since we can consider a line $DC$ as the fourth circle with zero curvature (infinite radius).
Thus, given that you've found values for $r_1,r_2$ ,
\begin{align} r_3&= \frac{r_1\,r_2}{r_1+r_2+2\sqrt{r_1\,r_2}} . \end{align}
The radius of the first circle is $\frac{1}{2}$, so its curvature is $k_1=2$. The radius of the second circle is $\frac{(\sqrt{2}-1)/2}{2\sqrt{2}}$, so its curvature is $k_2=\frac{4\sqrt{2}}{\sqrt{2}-1}$. The upper side of the square can be considered a circle with infinite radius and zero curvature. Thus, the Descartes theorem in this case reduces to $$k_4=k_1+k_2+2 \sqrt{k_1 k_2}$$ This gives
$$k_4=2+\frac{4\sqrt{2}}{\sqrt{2}-1}+2 \sqrt{ \frac{8\sqrt{2}}{\sqrt{2}-1} }\approx 26.109...$$
which corresponds to a radius of $$\frac{1}{26.109}\approx 0.038$$
The Circle in the Corner
In the corner of the square, the length of the red segment is $\frac1{\sqrt2}-\frac12$ (half the diagonal of the square minus the radius of the original circle). Because all these triangles are $45$-$45$-$90$, there is a ratio of $\sqrt2$ between sides. The length of the green segments is $1-\frac1{\sqrt2}$ and the length of the blue segment is $\sqrt2-1$. Thus, the perimeter of the green-green-blue triangle is $$ 2\left(\color{#090}{1-\frac1{\sqrt2}}\right)+\left(\color{#00F}{\sqrt2-1}\right)=1 $$ its area is $$ \frac12\left(\color{#090}{1-\frac1{\sqrt2}}\right)^2=\frac34-\frac1{\sqrt2} $$ The radius of the inscribed circle is $2$ times the area divided by the perimeter: $$ 2\,\frac{\frac34-\frac1{\sqrt2}}{1}=\frac{3-2\sqrt2}2 $$
The Radius of the Third Circle
To apply Descartes' Theorem, note that the bend (signed curvature) of the large circle, the small circle, and the top edge of the square are $2$, $6+4\sqrt2$, and $0$. Then Descartes' Theorem says $$ 2\left(2^2+\left(6+4\sqrt2\right)^2+0^2+\frac1{r^2}\right)=\left(2+\left(6+4\sqrt2\right)+0+\frac1r\right)^2\\ %144+96\sqrt2+\frac2{r^2}=96+64\sqrt2+\left(16+8\sqrt2\right)\frac1r+\frac1{r^2}\\ %\frac1{r^2}-\left(16+8\sqrt2\right)\frac1{r}+48+32\sqrt2=0\\ %\left(\frac1r-\left(8+4\sqrt2\right)\right)^2=48+32\sqrt2\\ %\frac1r-\left(8+4\sqrt2\right)=\pm\left(4+4\sqrt2\right)\\ %\frac1r=12+8\sqrt2\quad\text{or}\quad4\\ r=\color{#C00}{\frac{3-2\sqrt2}4}\quad\text{or}\quad\color{#090}{\frac14} $$ There are two solutions because there are two circles (shown below in red and green) tangent to the first two circles and the line containing the top of the square:
The radius of the circle in the question (the red circle) is $\color{#C00}{\frac{3-2\sqrt2}4}$.
The Center of the Third Circle
As shown in the corollary from this answer, the mean of the centers of the circles weighted by their bends equals their mean weighted by the square of their bends. Straight lines need special handling. Straight lines are essentially circles with a center at an infinite distance in a given direction. For the computation of centers:
$1$. the bend is $0$
$2$. the bend times the center is a unit vector perpendicular to the line
$\phantom{2\text{.}}$ and in the direction away from the other circles
$3$. the square of the bend times the center is $0$
For the red circle, solve $$ \tfrac{(0,1)+2(0,0)+\left(6+4\sqrt2\right)\left(1-\sqrt2,\sqrt2-1\right)+\color{#C00}{\left(12+8\sqrt2\right)(x,y)}}{2+\left(6+4\sqrt2\right)+\color{#C00}{\left(12+8\sqrt2\right)}} =\tfrac{2^2(0,0)+\left(6+4\sqrt2\right)^2\left(1-\sqrt2,\sqrt2-1\right)+\color{#C00}{\left(12+8\sqrt2\right)^2(x,y)}}{2^2+\left(6+4\sqrt2\right)^2+\color{#C00}{\left(12+8\sqrt2\right)^2}} $$ for $(x,y)$ to get $\color{#C00}{\left(\frac{\sqrt2-2}2,\frac{2\sqrt2-1}4\right)}$.
For the green circle, solve $$ \tfrac{(0,1)+2(0,0)+\left(6+4\sqrt2\right)\left(1-\sqrt2,\sqrt2-1\right)+\color{#090}{4(x,y)}}{2+\left(6+4\sqrt2\right)+\color{#090}{4}} =\tfrac{2^2(0,0)+\left(6+4\sqrt2\right)^2\left(1-\sqrt2,\sqrt2-1\right)+\color{#090}{4^2(x,y)}}{2^2+\left(6+4\sqrt2\right)^2+\color{#090}{4^2}} $$ for $(x,y)$ to get $\color{#090}{\left(-\frac{\sqrt2}2,\frac14\right)}$.
