Initial 49. Three circles with radii of $1, 2, 3$ are externally tangent to each other.
(1) Find the radius of the circle that is externally tangent to all three circles.
(Nanchang Liaoning Normal University High School Grade 1, Class 1, 116023)
(2) Find the radius of the circle that is internally tangent to all three circles.
(Li Xuewu, Tianjin Normal University Computer Department, 300074)
Solution: The side lengths of the triangle formed by the centers of the three circles are $3$, $4$, and $5$, so this triangle is a right triangle. Let the coordinates of its vertices be $(0,0)$, $(0,3)$, and $(4,0)$.
(1) Let the coordinates of the center of the required circle be $(u, v)$, and the radius be $x$. Then, using the tangency conditions, we get:
$$
\left\{\begin{array}{l}
u^{2}+v^{2}=(x+1)^{2}, \\
(u-4)^{2}+v^{2}=(x+3)^{2}, \\
u^{2}+(v-3)^{2}=(x+2)^{2} .
\end{array}\right.
$$

Eliminating $u, v$, we get
$$
23 x^{2}-132 x-36=0 \text{. }
$$

Solving for $x$, we get $x_{1}=\frac{6}{23}, x_{2}=-6$ (discard).
(2) Let the coordinates of the center of the required circle be $(s, t)$, and the radius be $y$. Then we have
$$
\left\{\begin{array}{l}
s^{2}+t^{2}=(y-1)^{2}, \\
(s-4)^{2}+t^{2}=(y-3)^{2}, \\
s^{2}+(t-3)^{2}=(y-2)^{2} .
\end{array}\right.
$$

Eliminating $s, t$, we get
$$
23 y^{2}-132 y-36=0 \text{. }
$$

Solving for $y$, we get $y_{1}=\frac{6}{23}, y_{2}=-\frac{6}{23}$ (discard).
(Note: The first part of this problem was provided by Nanchang, and only the answer $\frac{6}{23}$ was given.)