Question

In the given figure, from an external point P, a tangent PT and a line segment PAB drawn to a circle with centre O. ON is perpendicular on the chord AB.

Prove that

(i) PA × PB = PN² - AN²
(ii) PN² - AN² = OP² - OT²
(iii) PA PB = PT²​

Answer 1

Step-by-step explanation:

(i) PA . PB

=(PN – AN) (PN + BN)

=(PN – AN) (PN + AN) (As AN = BN)

=PN2 – AN2

(ii) PN2 – AN2

= (OP2 – ON2) – AN2 (As ON⊥PN)

= OP2 – (ON2 + AN2)

=OP2 – OA2 (As ON⊥AN)

= OP2 – OT2 (As OA = OT)

(iii) From (i) and (ii)

PA.PB = OP2 – OT2

= PT2 (As ∠OTP = 90°)