Question

AB is a chord of ⦿ (O,5) such that AB = 8. Tangents at A and B to the circle intersect in P. Find PA.

Answer 1

By drawing tangent to the circle.

and by using Pythagoras theorem.We got the following things as shown in the figure.

Answer 2

AB is a chord of circle of centre O(0,5) such that AB = 8. tangents at A and B to the circle intersect at P as shown in figure.

Let PR = x.
Since OP is a perpendicular bisector of AB,
AR = BR = 4
Consider ΔORA where ∠R = 90°,
By Pythagoras Theorem,
⇒ OA² = OR² + AR²
⇒ 5² = OR² + 4²
⇒ OR² = 25 – 16 = 9
∴ OR = 3
⇒ OP = PR + RO = x + 3 [ see figure]

Consider ΔARP where ∠R = 90°,
⇒ PA² = AR² + PR² [ by Pythagoras theorem]
⇒ PA² = 4² + x² = 16 + x² …...... (i)

Consider ΔOAP where ∠A = 90°,
⇒ PA² = OP² – OA² [by Pythagoras theorem]
= (x + 3)² – (5)²
= x² + 9 + 6x – 25
= x² + 6x – 16 …......... (ii)

now, From equations. (i) and (ii),
⇒ 16 + x²  = x²  + 6x – 16
⇒ 6x = 32
then, x = 16/3

from equation (i),
PA² = 16 + x² = 16 + (16/3)²
= 16(1 + 16/9) = 16 × 25/9
= 400/9
taking square root both sides,
so, PA = 20/3

therefore , PA = 20/3 = 6.67 m