Question

Do not spam his her account will be deleted as I am a moderator and offered 50 points

Answer 1

sᴏʟᴜᴛɪᴏɴ:

=> Draw ON _|_ AB

Therefore, AN = BN

Also we have PT = PA x PB. [ BY Tangent secant property].......(1)

Consider ∆ONA

$OA{}^{2} = \: {ON}^{2} + {AN}^{2}.....(2)$

Similarly inΔPTO

OP²= OT²+ PT²

That is PT² = OP² - OT²

= ON² + PN² - OA²  [Since OA = OT (radii)]

= ON+ PN²  – ON² – AN²

⇒ PT² = PN²– AN²----(3)

From (1) and (3), we get

PA × PB = PN²– AN²

PA × PB=T²

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Answer 2

Step-by-step explanation:

From figure:

(i)

Since ON ⊥ AB, N bisects AB. So, BN = AN.

PA.PB = (PN - AN)(PN + BN)

           = (PN - AN)(PN + AN)

           = PN² - AN²

(ii)

(a) In ΔONP, ∠ONP = 90°  (∵ ON ⊥ NP)

By Pythagoras theorem, PN² = OP² - ON²

(b) In ΔONA, ∠ONA = 90° (∵ON ⊥ NA)

By Pythagoras theorem, ON² + AN² = OA².

Now,

PN² - AN²

⇒ (OP² - ON²) - OA²

⇒ OP² - OA²

⇒ OP² - OT²

From (i) & (ii), we get

⇒ PA. PB = OP² - OT²

                = PT²

Therefore, PA.PB = PT²

Hope it helps!