Liberty : Assignment Paper - 2019 Std 10 Mathematics
32. AB is a chord of the circle with centre 0 and radius 5 such that AB = 8. Tangents at A and B to the circle
intersect in P. Find PA.
Answers
Answer:
Given AB is a chord of circle (O, 5) such that AB = 8.
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
Consider ΔARP where ∠R = 90°,
⇒ PA2 = AR2 + PR2
⇒ PA2 = 42 + x2 = 16 + x2 … (1)
Consider ΔOAP where ∠A = 90°,
⇒ PA2 = OP2 – OA2
= (x + 3)2 – (5)2
= x2 + 9 + 6x – 25
= x2 + 6x – 16 … (2)
From (1) and (2),
⇒ 16 + x2 = x2 + 6x – 16
⇒ 6x = 32
⇒ x =
From (1),
⇒ PA2 = 16 + x2
= 16 + 2
∴ PA =20/3
Answer:
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