A geometry problem by Adrian Gonzalez

Let $O$ be the center of the two circles and let $P$ be the point of tangency of $BC$ with the smaller circle.

Now let $x$ be the radius of the smaller circle, (and thus $3x$ the radius of the larger circle). Then triangles $ABC$ and $OPC$ are similar right triangles. So we thus have that

$\dfrac{AB}{AC} = \dfrac{OP}{OC} \Longrightarrow \dfrac{18}{6x} = \dfrac{x}{3x} \Longrightarrow x = 9$ .

The radius of the larger circle is thus $3x = \boxed{27}$ .

Let the point of tangency between chord $\overline{BC}$ and the smaller circle be point $P$ , and the center of the circle (it doesn't matter which one, as they are concentric) be point $O$ . Now, consider $\Delta OPC$ and $\Delta ABC$ . Note that $\angle OPC = \angle ABC = 90$ , and $\angle C = \angle C$ . Thus, the triangles are similar by AAA. Now, note that $\overline{AC}=2 \cdot \overline{OC}$ . Thus, the scale factor between the triangles is $2$ . Now, since $\overline{AB}=18$ , we have $\overline{OP}=\dfrac{\overline{AB}}{2} = \dfrac{18}{2}=9$ . Thus, the radius of the smaller circle is $9$ . Since the ratio of the smaller radius to the larger radius is $1:3$ , we see that the radius of the larger circle is $9 \cdot 3 = \boxed{27}$ .

Let O b center point and P be the point on small circle wher tangent touches. join line AB & OP both triangle are similar hence OP=1/2 AB AB=18 then OP=9 OP is radius of smaller circle hence smaler circle radius is 9 thn larger circel radius is 3xsmaller circle radius 3 x 9 27 Answer.