Question

If PAB is a secant to a circle intersecting the circle at A and B and PT is a tangent to the circle at T then prove that PT ² = PA . PB

Answer 1

Construction: Join AB 

Now, PA².PB² =(PM-AM)(PM+BM)   ...(As AM=BM) 

= (PM - AM) (PM + AM) 
= PM² - AM² 
= (OP² - OM²) - AM²
= OP² - (OM² + AM²) 
= OP² - OA² 

But, OA = OT           …(Radii ) 
So, PA x AB=PT²

Please see the other method too which I have attached ;)

Hope This Helps :)

Answer 2

Answer:

Join OL ⊥ AB

As O is the center of the circle, AL = BL

Now, PA × PB = (PL+AL)(PL-BL)

= (PL+AL)(PL-AL)

= PL² - AL²  ----- Eqn 1

Again, AL² = OA² - OL² (using Pythagoras theorem in ΔAOL)

∴ PL² - AL² = PL² - (OA² - OL²)

=  PL² - OA² + OL²

= (PL² + OL²) - OA²

= PO² - OA²  (using Pythagoras theorem in ΔPOL)

= PO² - OT² (As OA=OT, radii of the same circle)

= PT² (using Pythagoras theorem in ΔPOT)    ----- Eqn 2

∴ PA × PB = PT²