Question

In figure, from an external point la tangent
PT and a line segment PAB s drawn to
circle with centre O. ON is perpendicular
the chord AB. Prove that
PA.PB= OP²-OT²

Answer 1

Answer:

From theorem it is known that PT^2 = PA*PB ...........(1)

Now draw a line or just meet the point 'O' to 'T' than OPT become right angled triangle so in Triangle OPT.....

.

OP^2 = OT^2 + PT^2 , =>  OP^2 - OT^2 = PT^2 ....... (2)

from equan 1 and 2.....  PA.PB= OP²-OT² proved

Step-by-step explanation: