Question

In the given figure below, two concentric circles of radii a and b (a>b) are given. The chord AB of larger circle touches the smaller circle at C. The length of AB is
 A. $\sqrt {{a^2} - {b^2}} $
 B. $\sqrt {{a^2} + {b^2}} $
C. $2\sqrt {{a^2} - {b^2}} $
D. $2\sqrt {{a^2} + {b^2}} $

Answer 1

Hint-For solving such a question we just need to know the basic and proof of geometry as a perpendicular from the centre on a chord bisects the chord.

Complete step by step answer:
Given,
Radii are a and b where a>b,
In $\Delta $AOC,
$O{A^2} = O{C^2} + A{C^2}$
$
  A{C^2} = O{A^2} - O{C^2} \\
  A{C^2} = {a^2} - {b^2} \\
  AC = \sqrt {{a^2} - {b^2}} \\
$
We know that a perpendicular from centre of a circle to a chord bisects the chord, i.e. AB=2AC
AB=2$AC$=2$\sqrt {{a^2} - {b^2}} $
Here the answer is $2\sqrt {{a^2} - {b^2}} $

Note - For solving such a question we need to remember the proofs of geometry i.e. a perpendicular from centre of a circle to a chord bisect the chord. Proceeding like this you will get the right and answer to this problem.