​In the given figure, if PB = 20 cm and OA= 15 cm, then find the shortest distance between the circle and P.​

Correct option is D

Given:

PB = 20 cm

OA = 15 cm (radius of the circle)

O is the center of the circle

∠OAB and ∠OBA are right angles (tangents from point P)

Concept Used:

Pythagorean theorem to find OP

​$$OP^2 = OB^2 + PB^2$$​

Solution:

OB = radius = 15 cm

PB = 20 cm

​$$OP^2 = 15^2 + 20^2 = 225 + 400 = 625$$​

OP $$= \sqrt{625} = 25\ \text{cm}$$​

Shortest distance = OP − radius = 25 - 15 = 10 cm