It is given that AB = 5 and BC = 12

Using pythagoras theorem

$$\begin{gathered} {\mathrm{AC}}^2 = {\mathrm{AB}}^2 + {\mathrm{BC}}^2 \\ = 5^2 + {12}^2 \\ = 169 \\ \mathrm{Thus} \mathrm{AC} = 13 \end{gathered}$$

We know that  two tangents drwan to a circlefrom the same point that is exterior to

the  circle are of equal iengths.

Thus AM = AQ = a

Similarly  MB = BP = b  and  PC = CQ = c

We know AB = a+b = 5

BC = b+c = 12 and  AC = a+c = 13

Solving simultaneously we get  a = 3,  b =2,  c = 10

We also know that the tangent is perpendicular to the radius.

Thus OMBP is a square with side b

Hence the length of the radius of the circle inscribed in the right angled triangle is 2 cm.