15) Two circles have radii of $1 \mathrm{~m}$ and $3 \mathrm{~m}$, respectively. Knowing that there are two perpendicular lines, each tangent to both circles, what is the minimum possible distance between the two centers?
(A) $\sqrt{2} \mathrm{~m}$
(B) $2 \sqrt{2} \mathrm{~m}$,
(C) $(\sqrt{2}+\sqrt{5}) \mathrm{~m}$
(D) $2 \sqrt{5} \mathrm{~m}$,
(E) $4 \sqrt{2} \mathrm{~m}$
15. The answer is $\mathbf{( B )}$. Let's draw the two tangent lines to the two circles and the circle with radius $3 \mathrm{~m}$ positioned as in the figure, with center at $O$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d9e56cae5925cb0c0feag-4.jpg?height=511&width=522&top_left_y=2183&top_left_x=824)

The possible positions of the center of the circle with radius $1 \mathrm{~m}$ are the points $A, B, C, D$. Among the segments $O A, O B, O C$, and $O D$, the segment $A O$ has the minimum length; indeed, $O C$ is clearly longer than $O A$; but $O B$ (and $O D$ which is as long as $O B$) also has a length greater than $O A$ because $O A$ and $O B$ are the hypotenuses of the two right triangles $A H O$ and $B K O$ which have the two legs $A H$ and $B K$ of equal length, while the leg $O H$ is shorter than the leg $O K$. Therefore, the minimum distance between the centers of the two circles is equal to the length of $O A$; this can be calculated by applying the Pythagorean Theorem to the triangle

![](https://cdn.mathpix.com/cropped/2024_04_17_d9e56cae5925cb0c0feag-5.jpg?height=92&width=922&top_left_y=568&top_left_x=247)