Inscribed Tangent Circles in Right Triangle

In $\Delta ABC$, $AB = 18, BC = 24, AC = 30$. Draw a common internal tangent line $MN$ perpendicular to $BC$. This splits $\Delta ABC$ into a smaller right-angled triangle $\Delta MNC$ which is similar to $\Delta ABC$. In $\Delta MNC$, the second circle is the incircle of $\Delta MNC$. Ratio of sides of $\Delta MNC$ is $3:4:5$, so let $MN = 3x, NC = 4x, MC = 5x$. Inradius of $\Delta MNC$ is $r = \frac{3x + 4x - 5x}{2} = x$. For the first circle, its center is at distance $r$ from $AB$ and $BC$, and it touches the second circle, so $BN = 2r = 2x$. We have $BC = BN + NC = 2x + 4x = 6x = 24 \implies x = 4$. Thus, $r = 4\text{ cm}$.