Two concentric are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Step-by-Step Solution
Key Concept: For a chord of a circle, the perpendicular distance from the centre to the chord (let it be \(d\)) relates to the chord length \(l\) by \(l = 2\sqrt{R^{2}-d^{2}}\), where \(R\) is the radius of the circle. When the chord is tangent to an inner concentric circle, the distance \(d\) equals the radius of the inner circle.
1. Let \(O\) be the common centre of the two circles.
2. Radius of the larger circle \(R = 5\) cm and radius of the smaller circle \(r = 3\) cm.
3. The required chord \(AB\) of the larger circle touches the smaller circle; therefore the line \(AB\) is tangent to the inner circle at a point \(T\).
4. The radius \(OT\) is perpendicular to the tangent \(AB\). Hence the perpendicular distance from \(O\) to the chord \(AB\) is \(d = OT = r = 3\) cm.
5. In right triangle \(O M A\) (where \(M\) is the midpoint of \(AB\)), \(OM = d = 3\) cm and \(OA = R = 5\) cm.
6. Using Pythagoras theorem, \(AM = \sqrt{OA^{2} - OM^{2}} = \sqrt{5^{2} - 3^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4\) cm.
7. Since \(M\) is the midpoint, the whole chord length \(AB = 2 \times AM = 2 \times 4 = 8\) cm.
Correct Answer: 8 cm