Question

answer this please ✌️​

Answer 1

Answer:

If PAB is a secant to a circle intersecting it at A and B and PT is a tangent then PA.PB=PT

2

Here PAB is secant intersecting the circle with centre O at A and B and a tangent PT at T

We make construction that is OM⊥AB is drawn OA,OP,OT are joined.

Now PA=PM–AM

PB=PM+MB

As we know that AM=BM [perpendicular drawn from the centre of the circle to a chord is also a bisector of the chord]

PA.PB=(PM−AM).(PM+AM)

PA.PB=PM

2

−AM

2

Also OM⊥AB

we can apply Pythagoras theorem in ΔOMP

PM

2

=OP

2

−OM

2

Now apply Pythagoras theorem in ΔOMA

AM

2

=OA

2

−OM

2

now this values in the above equation so we get

PA.PB=PM

2

−AM

2

PA.PB=(OP

2

−OM

2

)−(OA

2

−OM

2

)

PA.PB=OP

2

−OM

2

−OA

2

+OM

2

PA.PB=OP

2

−OA

2

PA.PB=OP

2

−OT

2

as OA=OT (radii)

As radius is perpendicular to the tangent so this will form a right angled triangle

We can apply Pythagoras theorem in ΔOPT

We get PT

2

=OP

2

−OT

2

By putting this value in above equation we get PA.PB=PT

2