In the figure (i) given below, AB = 8 cm and M is mid-point of AB. Semicircles are drawn on AB, AM and MB as diameters. A circle with centre C touches all three semicircles as shown, find its radius.

Let x be the radius of the circle with centre C.

Since M is the mid-point of AB hence, AM = MB = 4 cm.

Two semicircles are thus drawn on AB with diameters as AM and MB.

Since radius = $\dfrac{\text{diameter}}{2}$.

Hence, radius of both the semicircles with diameters AM and MB = 2 cm.

From figure,

CM = MP - PC = (4 - x) cm.

In right angled triangle CMD,

$CD^2 = CM^2 + DM^2 \\[1em] (x + 2)^2 = (4 - x)^2 + 2^2 \\[1em] x^2 + 4 + 4x = 16 + x^2 - 8x + 4 \\[1em] x^2 - x^2 + 4x + 8x + 4 - 16 - 4 = 0 \\[1em] 12x - 16 = 0 \\[1em] 12x = 16 \\[1em] x = \dfrac{16}{12} \\[1em] x = \dfrac{4}{3} = 1\dfrac{1}{3} \text{ cm.}$

Hence, the radius of small circle = $1\dfrac{1}{3}$ cm.