A circular hoop and a rectangular frame are standing on the level ground as shown. The diagonal $$AB$$ is extended to meet the circular hoop at the highest point $$C$$. If $$AB = 18$$ cm, $$BC = 32$$ cm, the radius of the hoop (in cm) is ------.

Let's assume the circle touches the ground at O. Since C is the highest point on the circle, it must be diametrically opposite to O.

Now using the tangent-secant property

$$ AB\times AC = AO^2$$

$$ AO =\sqrt{18\times 50} = 30$$cm

Now, $$\triangle AOC$$ is a right-angled triangle

$$ AO^2 + CO^2 = AC^2$$

$$CO = \sqrt{50^2 - 30^2} = 40$$ cm.

Thus, the radius of the circle is $$20$$cm.