"Image by the authur" is misleading. The centers of the r-circle of the 5-circle should lie on the same line as their common tangent. [The problem is that the r-circle is drawn as an oval, not a circle. ]

Thus: (5-r)^2=x^2+r^2

where x is the distance along the x-axis from the center of the 5-circle to the tangent with the r-circle.

Likewise the centers of the 2.5-circle and the r-circle lie on the same line as their tangent. Thus:

(2.5+r)^2=(2.5+x)^2+r^2=2.5^2+5x+(5-r)^2

So: 5x=15r-25; x=3r-5.

And, thus: 25-10r+r^2=9r^2-30r+25+r^2; 9r^2=20r.

Since rcannot be zero, r=20/9. QED

Making the large radius 18 instead of 5 would make the solution more obvious;

The small triangle would be 6:8:10 and the large one 15:8:17.