# If PAB is a secant to a circle of Centre O intersecting the circle of A and B and PT is tangent segment, then prove that-

${\text{PA}} \times {\text{PB = P}}{{\text{T}}^2}$

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Hint – In this question first make a circle then draw a tangent and a secant from an external point of the circle and also, draw a perpendicular line from the Centre of the circle to the secant, then apply Pythagoras theorem .

Given Data, PAB is a secant, Centre of circle O, PT is a tangent

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

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

Now PA = PM – AM

PB = PM + MB

As we know AM = BM (perpendicular drawn from the Centre of the circle to a chord is also a bisector of chord)

PA.PB = (PM – AM) × (PM + AM) = ${\text{P}}{{\text{M}}^2} - {\text{A}}{{\text{M}}^2}{\text{ }} \to {\text{ Equation 1}}$

Also from the diagram OM ⊥ AB

We can apply Pythagoras theorem in ∆OMP

${\text{P}}{{\text{M}}^2} = {\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{M}}^2}$

We can apply Pythagoras theorem in ∆OMA

${\text{A}}{{\text{M}}^2} = {\text{O}}{{\text{A}}^2} - {\text{O}}{{\text{M}}^2}$

Put these values in Equation 1, we get

PA.PB = ${\text{P}}{{\text{M}}^2} - {\text{A}}{{\text{M}}^2}$

PA.PB = (${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{M}}^2}$) – (${\text{O}}{{\text{A}}^2} - {\text{O}}{{\text{M}}^2}$)

PA.PB = ${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{A}}^2}$

As OA = OT (radii)

PA.PB = ${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{T}}^2}{\text{ }} \to {\text{ Equation 2}}$

As radius is perpendicular to the tangent so this will form a right angled triangle, we can apply Pythagoras theorem in ∆OPT

We get ${\text{P}}{{\text{T}}^2}{\text{ = O}}{{\text{P}}^2} - {\text{O}}{{\text{T}}^2}$

By putting this in the Equation 2, we get

PA.PB = ${\text{P}}{{\text{T}}^2}$

Hence Proved.

Note – In order to solve this type of questions, draw an appropriate figure and construct any other additional elements using known geometrical properties. Pythagoras theorem can be used if we construct a right angled triangle. Properties of straight lines are very important in order to approach the answer or proof.

__Complete step-by-step solution -__Given Data, PAB is a secant, Centre of circle O, PT is a tangent

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

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

Now PA = PM – AM

PB = PM + MB

As we know AM = BM (perpendicular drawn from the Centre of the circle to a chord is also a bisector of chord)

PA.PB = (PM – AM) × (PM + AM) = ${\text{P}}{{\text{M}}^2} - {\text{A}}{{\text{M}}^2}{\text{ }} \to {\text{ Equation 1}}$

Also from the diagram OM ⊥ AB

We can apply Pythagoras theorem in ∆OMP

${\text{P}}{{\text{M}}^2} = {\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{M}}^2}$

We can apply Pythagoras theorem in ∆OMA

${\text{A}}{{\text{M}}^2} = {\text{O}}{{\text{A}}^2} - {\text{O}}{{\text{M}}^2}$

Put these values in Equation 1, we get

PA.PB = ${\text{P}}{{\text{M}}^2} - {\text{A}}{{\text{M}}^2}$

PA.PB = (${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{M}}^2}$) – (${\text{O}}{{\text{A}}^2} - {\text{O}}{{\text{M}}^2}$)

PA.PB = ${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{A}}^2}$

As OA = OT (radii)

PA.PB = ${\text{O}}{{\text{P}}^2} - {\text{O}}{{\text{T}}^2}{\text{ }} \to {\text{ Equation 2}}$

As radius is perpendicular to the tangent so this will form a right angled triangle, we can apply Pythagoras theorem in ∆OPT

We get ${\text{P}}{{\text{T}}^2}{\text{ = O}}{{\text{P}}^2} - {\text{O}}{{\text{T}}^2}$

By putting this in the Equation 2, we get

PA.PB = ${\text{P}}{{\text{T}}^2}$

Hence Proved.

Note – In order to solve this type of questions, draw an appropriate figure and construct any other additional elements using known geometrical properties. Pythagoras theorem can be used if we construct a right angled triangle. Properties of straight lines are very important in order to approach the answer or proof.

Last updated date: 20th Sep 2023

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