Two circles $C_{1}$ and $C_{2}$ touch each other externally and the line $l$ is a common tangent. The line $m$ is parallel to $l$ and touches the two circles $C_{1}$ and $C_{3}$. The three circles are mutually tangent. If the radius of $C_{2}$ is 9 and the radius of $C_{3}$ is 4 , what is the radius of $C_{1}$ ?

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-146.jpg?height=252&width=460&top_left_y=209&top_left_x=1277)
(A) 10.4
(B) 11
(C) $8 \sqrt{2}$
(D) 12
(E) $7 \sqrt{3}$
Two circles $C_{1}$ and $C_{2}$ touch each other externally and the line $l$ is a common tangent. The line $m$ is parallel to $l$ and touches the two circles $C_{1}$ and $C_{3}$. The three circles are mutually tangent. If the radius of $C_{2}$ is 9 and the radius of $C_{3}$ is 4 , what is the radius of $C_{1}$ ?
(A) 10.4
(B) 11
(C) $8 \sqrt{2}$
(D) 12
(E) $7 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-247.jpg?height=284&width=504&top_left_y=709&top_left_x=1401)

## Solution

We start by joining the centres of the circles to form $\Delta C_{1} C_{2} C_{3}$. (The lines joining the centres pass through the corresponding points of tangency.)

Secondly, we construct the rectangle $A B C_{2} D$ as shown in the diagram. If the radius of the circle with centre $C_{1}$ is $r$ we see that: $C_{1} C_{2}=r+9, C_{1} C_{3}=r+4$ and $C_{2} C_{3}=13$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-247.jpg?height=288&width=504&top_left_y=1233&top_left_x=1401)

We now label lengths on the rectangle in the way noted in the diagram.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-247.jpg?height=325&width=530&top_left_y=1583&top_left_x=1383)

To understand this labelling, look for example at $C_{1} D$. The radius of the large circle is $r$ and the radius of the circle with centre $C_{2}$ is 9 . The length $C_{1} D$ is then $r-9$.

This same kind of reasoning can be applied to both $C_{1} A$ and $B C_{2}$.

Using Pythagoras we can now derive the following:

In $\triangle A C_{3} C_{1}$,

$$
\begin{aligned}
C_{3} A^{2} & =(r+4)^{2}-(r-4)^{2} \\
& =16 r
\end{aligned}
$$

Therefore $C_{3} A=4 \sqrt{r}$.

In $\triangle D C_{1} C_{2}$,

$$
\begin{aligned}
\left(D C_{2}\right)^{2} & =(r+9)^{2}-(r-9)^{2} \\
& =36 r
\end{aligned}
$$

Therefore $D C_{2}=6 \sqrt{r}$.

In $\triangle B C_{3} C_{2}, \quad\left(C_{3} B\right)^{2}=13^{2}-(2 r-13)^{2}$

$$
=-4 r^{2}+52 r
$$

Therefore $C_{3} B=\sqrt{-4 r^{2}+52 r}$.

In a rectangle opposite sides are equal, so:

$$
\begin{aligned}
& D C_{2}=C_{3} A+C_{3} B \\
& \text { or, } \quad 6 \sqrt{r}=4 \sqrt{r}+\sqrt{-4 r^{2}+52 r} \\
& 2 \sqrt{r}=\sqrt{-4 r^{2}+52 r}
\end{aligned}
$$

Squaring gives, $4 r=-4 r^{2}+52 r$

$$
\begin{aligned}
& 4 r^{2}-48 r=0 \\
& 4 r(r-12)=0
\end{aligned}
$$

Therefore $r=0$ or $r=12$.

Since $r>0, r=12$.