From an external point $P$ , a tangent $P T$ and a line segment $P A B$ are drawn to a circle with center $O$ . Prove that $P A . P B = P T^{2}$

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Solution

We have,

From an external point P, a tangent PT and a line segment PAB are drawn to a circle with centre O.

Prove that:- $P A . P B = P T^{2}$

Proof:-

According to figure,

Draw $O N ⊥ A B$

Therefore, $A N = B N$

Also,

We have,

$P T = P A \times P B (By tangent secant property) . . . . . . (1)$

$I n Δ O N A,$

$O A^{2} = O N^{2} + A N^{2} . . . . . . (2)$

Similarly, In $Δ P T O,$

$O P^{2} = O T^{2} + P T^{2}$

$P T^{2} = O P^{2} - O T^{2}$

$\Rightarrow O N^{2} + P N^{2} - O A^{2} (S i n c e O A = O T)$

$\Rightarrow O N^{2} + P N^{2} - O N^{2} - A N^{2}$

$\Rightarrow P T^{2} = P N^{2} - A N^{2} . . . . . . (3)$

From equation (1) and (3) to,

$P A \times P B = P N^{2} - A N^{2}$

$P A \times P B = P T^{2}$

Hence, proved.

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P

P T

P A B

O

O N

A B

(i) P A . P B = {P A}^{2} - {A N}^{2}

(i i) {P N}^{2} - {A N}^{2} = {O P}^{2} - {O T}^{2}

(i i i) P A . P B = {P T}^{2}

∠

= 2 ∠

Two tangents PA and PB are drawn to a circle with center O from an external point P. Prove that $∠ A P B = 2 ∠ O A B$ [3 MARKS]

P A B

O

A

B

P T

P A \times P B = P T^{2}