In the adjoining figure, four successively touching circles are placed in the interior of $$\angle AOB$$. The first (smallest) has a radius 7 cm . The third circle has a radius 28 cm . Then the radius of the largest circle (in cm ) is

Consider 

$$\triangle OCE, \triangle ODF$$

$$ \angle OCE = \angle ODF = 90\degree$$ (Angle made by radius with the tangent)

$$\angle COE = \angle DOF$$

Thus, by AA similarity criterion, the triangles are similar to each other. We can extend this to all the circles to find that the triangles made in such a way by the line passing through their centres, the radius and the tangent are all similar.

Let $$OE = x$$

Thus, using similarity we can write

$$ \dfrac{x}{r_1} = \dfrac{x + r_1+r_2}{r_2} $$

We can further extend this to all four circles to write

$$ \dfrac{x}{r_1} = \dfrac{x + r_1+r_2}{r_2} = \dfrac{x+r_1+r_2+r_3}{r_3} = \dfrac{x+r_1+r_2+r_3+r_4}{r_4}$$

Let the common value be $$k$$.

Then,

$$\dfrac{x}{r_1}=k$$

$$x=kr_1.$$

Also,

$$\dfrac{x+r_1}{r_2}=k$$

$$x+r_1=kr_2.$$

Since $$x=kr_1$$,

$$kr_1+r_1=kr_2$$

$$r_1(k+1)=kr_2$$

Therefore,

$$\dfrac{r_2}{r_1}=\dfrac{k+1}{k}.$$

Now consider the next pair:

$$\dfrac{x+r_1+r_2}{r_3}=k.$$

Since $$x+r_1=kr_2$$,

$$x+r_1+r_2=kr_2+r_2$$

$$=r_2(k+1).$$

Therefore,

$$kr_3=r_2(k+1)$$

and hence

$$\dfrac{r_3}{r_2}=\dfrac{k+1}{k}.$$

Similarly,

$$\dfrac{r_4}{r_3}=\dfrac{k+1}{k}.$$

Thus,

$$\boxed{\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}=\dfrac{r_4}{r_3}}$$

Therefore, the radii form a geometric progression.

$$\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}=\dfrac{r_4}{r_3}$$

Let this common ratio be $$k$$.

Thus, the radii form a geometric progression:

$$r_1,\ r_2,\ r_3,\ r_4$$

We are given

$$r_1=7$$ and $$r_3=28$$

Therefore,

$$r_3=r_1k^2$$

$$28=7k^2$$

$$\implies k^2=4$$

Since the radii are increasing, $$k=2$$.

Therefore,

$$r_4=r_3k$$

$$r_4=28(2)$$

$${r_4=56}$$