Given, a circle is of radius16 cm touching a circle of radius r at an externally and

and the circles with radius r touches a circle with radius 4 cm at B externally.

CD is a common tangent to the 3 circles.

From the figure,

The horizental line is the equivalent of a circle of curvature 0.

Given, circle have curvatures 0, 1, 1/4 and 1/r

Now, apply Descartes circle formula, we get

      2(0² + 1² + 1/4² + 1/r² ) = (0 + 1 + 1/4 + 1/r)²

=> 2(1 + 1/16 + 1/r² ) = (5/4 + 1/r)²

=> 2(17/16 + 1/r² ) = 25/16 + 5/2r + 1/r²

=> 34/16 + 2/r² = 25/16 + 5/2r + 1/r²

=> 34/16 + 2/r² - 25/16 - 5/2r - 1/r² = 0

=> 1/r² - 5/2r + 9/16 = 0

=> 16/r² - 40/2r + 9 = 0

=> (4/r - 9)*(4/r - 1) = 0

=> r = 4, 4/9

Since r = 4 is not possible

So, r = 4/9

Hence, the radius fo the circle is 4/9 cm