Question

Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

• 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

The correct option is C

$2\sqrt{a^2-b^2}$

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then OC $\perp$ AB
Let OA = a and OC = b.

Since OC $\perp$ AB, OC bisects AB
[ $\because$ perpendicular from the centre to a chord bisects the chord].

In right $\Delta$ ACO, we have

$O{A}^2=O{C}^2+A{C}^2$ [by Pythagoras' theorem]

$\Rightarrow AC=\sqrt{O{A}^2-O{C}^2}=\sqrt{a^2-b^2}$
$\therefore AB=2AC=2\sqrt{a^2-b^2}$ [ $\because$ C is the midpoint of AB]

i.e., Length of the chord $AB=2\sqrt{a^2-b^2}$