$C, C_1, C_2, C_3$ are three circles of radii $5, 3, 2, x$ units respectively. $C_1$ and $C_2$ touch each other externally and $C$ internally. $C_3$ touches $C_1$ and $C_2$ externally and $C$ internally.
Find the value of $x$.
I've figured out that the diagram looks like this. And I have a hunch that the centre of $C_3$ will be directly above the intersection point of $C_1$ and $C_2$. I've also tried constructing right-angled triangles and applying the pythagoras theorem.
(see picture below)
Let us denote, on homogeneity grounds, $C_0=C$ and by $C_{ij}$ the contact point of circle $C_i$ with circle $C_j$.
A common feature is that points $C_i,C_j$ and $C_{ij}$ are aligned. More precisely, we are interested by the following alignements:
$C_1,C_{13},C_3$ are aligned in this order, thus $C_1C_3=3+x$ (external contact).
$C_2,C_{23},C_3$ are aligned in this order, thus $C_2C_3=2+x$ (external contact).
$C_0,C_3,C_{03}$ are aligned in this order, thus $C_2C_3=5-x$ (internal contact).
Let $(x_3,y_3)$ be the coordinates of $C_3$.
We can write the following system of 3 equations with 3 unknowns $x,x_3,y_3$:
$$\begin{cases}\text{square of length} \ C_1C_3: & \ \ (x_3-3)^2+y_3^2&=&(3+x)^2 & \ \ (a)\\\text{square of length} \ C_2C_3: & \ \ (x_3-8)^2+y_3^2&=&(2+x)^2&\ \ (b)\\ \text{square of length} \ C_0C_3: & \ \ (x_3-5)^2+y_3^2&=&(5-x)^2&\ \ (c) \\ \end{cases}$$
Expanding and subtracting (a) from (b) and (b) from (c) we obtain a linear system with $x$ and $x_3$ as unknowns, giving
together with $x_3=120/19$ and, after plugging these results into (a): $y_3=60/19$.