In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm.

AB is the diameter of the circle with centre O, which bisects the chord CD at point E.
Given: CE = ED = 8 cm and EB = 4 cm
Join OC.

Let OC = OB = r cm   (Radii of a circle)
Then OE = (r − 4) cm
Now, in right angled ΔOEC, we have:

$O C^{2}=O E^{2}+E C^{2}$        (Pythagoras theorem)

$\Rightarrow r^{2}=(r-4)^{2}+8^{2}$

$\Rightarrow r^{2}=r^{2}-8 r+16+64$

$\Rightarrow r^{2}-r^{2}+8 r=80$

$\Rightarrow 8 r=80$

$\Rightarrow r=\left(\frac{80}{8}\right) \mathrm{cm}=10 \mathrm{~cm}$

⇒ r = 10 cm
Hence, the required radius of the circle is 10 cm.