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Answer :

Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P. Join OP.

∵ OP is the radius of the smaller circle and AB is tangent to this circle at P,

∴ \( OP \ ⊥ \ AB \).

We know that, the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

So, \( OP \ ⊥ \ AB \) and \( AP \ = \ BP \).

In \( ∆ \ APO \), we have

\( OA^2 \ = \ AP^2 \ + \ OP^2 \ => \ 5^2 \ = \ AP^2 \ + \ 3^2 \)

\( AP^2 \ = \ 5^2 \ - \ 3^2 \ = \ 25 \ - \ 9 \ = \ 16 \)

\( AP \ = \ 4 \)

Now, \( AB \ = \ 2AP \) [∵ \( AP \ = \ BP \) ]

\( AB = \ 2 \ × \ 4 \ = \ 8 \)

∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

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- If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then \( ∠ \ POA \) is equal to (A) \( 50° \) (B) \( 60° \) (C) \( 70° \) (D) \( 80° \)
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