Three [circles](https://artofproblemsolving.com/wiki/index.php/Circle) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) $s$ are drawn in the first [quadrant](https://artofproblemsolving.com/wiki/index.php/Quadrant) of the $xy$-[plane](https://artofproblemsolving.com/wiki/index.php/Plane). The first circle is tangent to both axes, the second is [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_(geometry)) to the first circle and the $x$-axis, and the third is tangent to the first circle and the $y$-axis. A circle of radius $r > s$ is tangent to both axes and to the second and third circles. What is $r/s$?

$(\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10$
Solution 1
[2005 12A AMC-16b.png](https://artofproblemsolving.com/wiki/index.php/File:2005_12A_AMC-16b.png)
Set $s =1$ so that we only have to find $r$. Draw the segment between the center of the third circle and the large circle; this has length $r+1$. We then draw the [radius](https://artofproblemsolving.com/wiki/index.php/Radius) of the large circle that is perpendicular to the [x-axis](https://artofproblemsolving.com/wiki/index.php/X-axis), and draw the perpendicular from this radius to the center of the third circle. This gives us a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) with legs $r-3,r-1$ and [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) $r+1$. The [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) yields:

$(r-3)^2 + (r-1)^2 = (r+1)^2$$r^2 - 10r + 9 = 0$$r = 1, 9$
Quite obviously $r > 1$, so $r = 9 \