Question

plz solve it quickly ​

Answer 1

O is the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.

Draw QP ⊥ RP at point P, such that point Q lies on the circle.

∠OPR = 90° (radius ⊥ tangent)

Also, ∠QPR = 90° (Given)

∴ ∠OPR = ∠QPR

Now, above case is possible only when centre O lies on the line QP.

Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Answer 2

Step-by-step explanation:

O is the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.

Draw QP I RP at point P, such that point

Q lies on the circle.

ZOPR = 90° (radius I tangent)

Also, ZQPR = 90° (Given)

: SOPR = 2QPR

Now, above case is possible only when centre O lies on the line QP.