Question:

Two concentric circles are of radii $8\ \text{cm}$ and $5\ \text{cm}$. Find the length of the chord of the larger circle which touches (is tangent to) the smaller circle.
 

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For a chord tangent to an inner concentric circle, the distance from the common centre to the chord equals the inner radius. Use $\,\text{chord}=2\sqrt{R^{2}-d^{2}}$.
Updated On: Sep 9, 2025
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Solution and Explanation


Let $O$ be the common centre, $R=8\ \text{cm}$ the radius of the larger circle, and $r=5\ \text{cm}$ the radius of the smaller circle.
If a chord $AB$ of the larger circle touches the smaller circle, then the perpendicular distance from $O$ to the chord equals the smaller radius: $d=OA_\perp = r=5\ \text{cm}$.
For a circle, the length of a chord at distance $d$ from the centre is \[ AB=2\sqrt{R^{2}-d^{2}}. \] Hence, \[ AB=2\sqrt{8^{2}-5^{2}}=2\sqrt{64-25}=2\sqrt{39}\ \text{cm}\approx 12.49\ \text{cm}. \] \[ \boxed{\,\text{Chord length}=2\sqrt{39}\ \text{cm}\ \ (\approx 12.49\ \text{cm})\,} \]
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