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_ac36362783317e0251fdg-136.jpg?height=608&width=563&top_left_y=539&top_left_x=1258)
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_33a8a6559d675f59baa2g-270.jpg?height=567&width=510&top_left_y=275&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_33a8a6559d675f59baa2g-270.jpg?height=710&width=683&top_left_y=949&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_33a8a6559d675f59baa2g-270.jpg?height=241&width=306&top_left_y=1701&top_left_x=1430)

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_33a8a6559d675f59baa2g-271.jpg?height=710&width=613&top_left_y=279&top_left_x=1279)