P is a point outside a circle and is 26 cm away from its centre. A secant PAB drawn from intersects the circle at points A and B such that PB = 32 cm and PA= 18 cm. The radius of the circle (in cm)is:

From the given question we draw the diagram 

OA is Radius of circle  PA = 18cm and PB= 32cm

then AB= 32-18 = 14

then BD=AD = 7 cm

In the $$\triangle ODP,$$

                          $$ (OD)^2 = (OP)^2 - (DP)^2 $$

                          $$ \Rightarrow (OD)^2 = (26)^2-(25)^2 $$

                         $$\Rightarrow( OD)^2 = 676-625 $$

                       $$\Rightarrow OD = \sqrt{51} $$

then In $$\triangle OAD, $$

   $$ r^2 = (OD)^2 + (AD)^2 $$

  $$\Rightarrow r^2 = (\sqrt {51})^2 + (7)^2 $$

 $$\Rightarrow r^2 =  51+49 $$

$$\Rightarrow r^2 = 100 $$

$$\Rightarrow r = 10 $$cm Ans