The circle with centre $A$ has radius 3 and is tangent to both the positive $x$-axis and positive $y$-axis, as shown. Also, the circle with centre $B$ has radius 1 and is tangent to both the positive $x$-axis and the circle with centre $A$. The line $L$ is tangent to both circles. The $y$-intercept of line $L$ is
(A) $3+6 \sqrt{3}$
(B) $10+3 \sqrt{2}$
(C) $8 \sqrt{3}$
(D) $10+2 \sqrt{3}$
(E) $9+3 \sqrt{3}$
The circle with centre $A$ has radius 3 and is tangent to both the positive $x$-axis and positive $y$-axis, as shown. Also, the circle with centre $B$ has radius 1 and is tangent to both the positive $x$-axis and the circle with centre $A$. The line $L$ is tangent to both circles. The $y$-intercept of line $L$ is
(A) $3+6 \sqrt{3}$
(B) $10+3 \sqrt{2}$
(C) $8 \sqrt{3}$
(D) $10+2 \sqrt{3}$
(E) $9+3 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-250.jpg?height=564&width=510&top_left_y=995&top_left_x=1320)

## Solution

We start by drawing a line from point $C$ that will pass through $A$ and $B$. From $A$ and $B$, we drop perpendiculars to the points of tangency on the $x$-axis and label these points as $E$ and $F$ as shown. We also drop a perpendicular from $A$ to the $y$-axis which makes $A H=A E=3$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-250.jpg?height=705&width=683&top_left_y=1664&top_left_x=1233)

Extracting $\triangle C A E$ from the diagram and labelling with the given information we would have the following noted in the diagram.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-251.jpg?height=234&width=304&top_left_y=276&top_left_x=1428)

If we represent the distance from $C$ to $B$ as $x$ and recognize that $\triangle C B F$ is similar to $\triangle C A E$,

$$
\begin{aligned}
& \frac{x}{1}=\frac{x+4}{3} \\
& x=2 \\
& F C^{2}=2^{2}-1^{2}=3 \\
& \quad F C=\sqrt{3},(F C>0)
\end{aligned}
$$

In $\triangle C B F, F C^{2}=2^{2}-1^{2}=3$

This implies that $\angle B C F=30^{\circ}$ and $\angle O C D=60^{\circ}$. Therefore $E F=2 \sqrt{3}$, from similar triangles again.

This now gives us the diagram shown.

Thus, $d=\sqrt{3}(3+3 \sqrt{3})$

$$
=3 \sqrt{3}+9
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-251.jpg?height=707&width=604&top_left_y=1067&top_left_x=1292)