AB is the diameter of a circle. The chord CD is perpendicular to AB intersecting it at P. If CP=2 and PB=1, the radius of the circle is

Correct option is B

Given:

• CP = 2 units
• PB = 1 unit
• AB is the diameter of the circle
• CD is perpendicular to AB and intersects AB at point P
• O is the center of the circle, so AB = 2r and OB = r

Concept:

• Pythagoras' theorem states h² = p² + b² where h is the hypotenuse, p is perpendicular, and b is the base of a right-angled triangle.
• (a + b)(a - b) = a² - b². 

Solution:

Given CP = 2 and PB = 1

Let the radius of the circle be r.

=> OC = OB = r.

Consider triangle OCP.

CP = 2 (given)

OC = r (radius)

OP = OB - PB.

OP = r - 1

Δ OCP is a right-angled triangle at P.

∴ OC² = OP² + CP² (Pythagoras theorem)

=> r² = (r - 1)² + 2².

=> r² - (r - 1)² = 4.

=>(r² -( r² + 1 - 2r)) = 4.

=>(2r - 1) = 4.

=> 2r = 4 + 1

=> 2r = 5.

=> r = 2.5

The correct answer is option B.