4. Three circles $C_{1}, C_{2}, C_{3}$ with radii of lengths 1, 2, and 3 respectively touch each other externally. In the enclosed area lies a small circle $C_{4}$ that touches each of the three given circles externally.

Calculate the length of the radius of $C_{4}$.
4. Three circles $C_{1}, C_{2}, C_{3}$ with radii of lengths 1, 2, and 3 respectively touch each other externally. In the enclosed area lies a small circle $C_{4}$ that touches each of the three given circles externally. Calculate the length of the radius of $C_{4}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_67d91aa7ad73b4bdc7fdg-2.jpg?height=353&width=307&top_left_y=119&top_left_x=1668)

Solution: Let the centers of $C_{1}, C_{2}, C_{3}$ be $A, B, C$ respectively. The triangle formed by $A, B$, and $C$ has sides $|A B|=3,|A C|=4$, and $|B C|=5$, and is thus a right triangle. Choose a coordinate system with the right angle $A$ as the origin and the other two centers on the axes: $B=(3,0)$ and $C=(0,4)$. Let the radius and center of $C_{4}$ be $r$ and $M=(x, y)$ respectively. Then we have: $|A M|=r+1,|B M|=r+2$, and $|C M|=r+3$. This leads to the following three equations with three unknowns:

$$
\begin{aligned}
& r^{2}+2 r+1=|A M|^{2}=x^{2}+y^{2} \\
& r^{2}+4 r+4=|B M|^{2}=(3-x)^{2}+y^{2}=x^{2}-6 x+9+y^{2} \\
& r^{2}+6 r+9=|C M|^{2}=x^{2}+(4-y)^{2}=x^{2}+y^{2}-8 y+16
\end{aligned}
$$

The difference between (1) and (2) gives $6 x-9=-2 r-3$, so $6 x=6-2 r$, thus $x=\frac{3-r}{3}$. The difference between (1) and (3) gives $8 y-16=-4 r-8$, so $8 y=8-4 r$, thus $y=\frac{2-r}{2}$. Substituting into (1) gives the equation $r^{2}+2 r+1=x^{2}+y^{2}=\frac{(3-r)^{2}}{9}+\frac{(2-r)^{2}}{4}=\frac{9-6 r+r^{2}}{9}+\frac{4-4 r+r^{2}}{4}$, so

$$
\frac{23}{36} r^{2}+\frac{11}{3} r-1=23\left(\frac{r}{6}\right)^{2}+22\left(\frac{r}{6}\right)-1=0 .
$$

With $p=\frac{r}{6}$, this equation becomes $0=23 p^{2}+22 p-1=$ $(23 p-1)(p+1)$, so $\frac{r}{6}=p=-1$ (but this gives $r=-6<0$, so this solution is invalid) or $23 \cdot \frac{r}{6}=23 p=1$, so $r=\frac{6}{23}$.

(From this, we can also deduce that $M=(x, y)=$ $\left.\left(\frac{21}{23}, \frac{20}{23}\right).\right)$