Find the Radius of a Circle Tangent to a Triangle

In this geometry problem, a triangle is tangent to a circle at two points. We are given specific lengths and must determine the radius of the circle using right triangle relationships and algebra.

Problem

In the figure below, triangle \(ABC\) is tangent to a circle with center \(O\) at two points. The lengths are:

\[ AM = 6 \text{ cm}, \quad BC = 18 \text{ cm} \]

Find the radius \(r\) of the circle.

Triangle tangent to circle geometry problem diagram

Solution

Let \(B\) and \(N\) be the points of tangency. Since a radius is perpendicular to the tangent at the point of tangency, several right triangles are formed.

Triangle tangent to circle geometry solution diagram

Step 1: Apply the Pythagorean Theorem to Triangle \(AON\)

Let \(r\) be the radius of the circle. Then:

\[ AN^2 + r^2 = (r + 6)^2 \]

Simplifying:

\[ AN^2 = 36 + 12r \] \[ AN = \sqrt{36 + 12r} \]

Step 2: Use Congruent Right Triangles

Triangles \(ONC\) and \(OBC\) are right triangles with \(ON = OB = r\). Therefore, they are congruent and:

\[ NC = BC = 18 \]

Step 3: Apply the Pythagorean Theorem to Triangle \(ABC\)

\[ (6 + 2r)^2 + 18^2 = (18 + AN)^2 \]

Step 4: Expand and Simplify

After expanding and simplifying: \[ 4r^2 + 12r = 36AN \] Substitute \(AN = \sqrt{36 + 12r}\): \[ r^2 + 3r = 9\sqrt{36 + 12r} \]

Step 5: Square Both Sides

\[ (r^2 + 3r)^2 = 81(36 + 12r) \] Expanding: \[ r^4 + 6r^3 + 9r^2 - 972r - 2916 = 0 \]

Step 6: Solve the Equation

The equation has two real solutions. Only the positive solution is valid for a radius:

\[ \boxed{r = 9 \text{ cm}} \]

Final Answer

\[ \boxed{9 \text{ cm}} \]