This is not an elementary problem; in fact, it's known as the Decartes' circle theorem; see, for example,
http://en.wikipedia.org/wiki/Descartes%27_theorem

In their formula (2), which I've attached, the k's represent the reciprocals of the radii. Thus $\frac{1}{k_4}$ is the RADIUS of the circle you are after; use $k_1 = \frac{1}{25}, k_2 = \frac{1}{225}, \mbox{ and } k_3 = \frac{1}{30},$ and use the MINUS sign. (To understand the formula, practice with the radii of the three starting tangent circles all equal to 1, in which case you can find the radius, which is about 2.15 -- and diameter 4.30 about -- of the outer circumscribing circle using elementary geometry.)

\[k_4 = k_1 + k_2 + k_3 \pm 2 \sqrt{k_1k_2 + k_2k_3 + k_3k_1}.\]

A very nice proof that the formula works can be found on pages 14-15 of INTRODUCTION TO GEOMETRY by H.S.M. Coxeter, but there are dozens of proofs that you can find in books and on the internet.

Chris