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.
Answer: Option C
:
C
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 ⊥ AB
LetOA = a and OC = b.
SinceOC ⊥ AB, OC bisects AB
[ ∵ perpendicular from the centre to a chord bisects the chord].
In right Δ ACO, we have
OA2=OC2+AC2 [by Pythagoras' theorem]
⇒AC=√OA2−OC2=√a2−b2
∴AB=2AC=2√a2−b2 [ ∵ C is the midpoint of AB]
i.e., Length of the chord AB=2√a2−b2
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:
C
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 ⊥ AB
LetOA = a and OC = b.
SinceOC ⊥ AB, OC bisects AB
[ ∵ perpendicular from the centre to a chord bisects the chord].
In right Δ ACO, we have
OA2=OC2+AC2 [by Pythagoras' theorem]
⇒AC=√OA2−OC2=√a2−b2
∴AB=2AC=2√a2−b2 [ ∵ C is the midpoint of AB]
i.e., Length of the chord AB=2√a2−b2
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