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Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact. Hence, find the length of the chord of the larger circle which touches the smaller circle, if the diameter of concentric circles are $10$ cm and $8$ cm respectively.
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For correct Figure: (1 Mark)
nLet O be the centre of the circles and AB be the chord of larger circle, touching the smaller circle at C.
n$OC \perp AB$ [Tangent to a circle $\perp$ the radius through the point of contact] (1 Mark)
n$\therefore AC = CB$ [$\perp$ drawn from centre of a circle to a chord bisects the chord] (1 Mark)
n$OC = 4$cm and $OB = 5$ cm (1/2 Mark)
n$\therefore$ by Pythagoras theorem in $\Delta BOC$,
n$BC = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ cm (1/2 Mark)
n$\therefore AB = 2BC = 2 \times 3 = 6$ cm (1 Mark)
nLet O be the centre of the circles and AB be the chord of larger circle, touching the smaller circle at C.
n$OC \perp AB$ [Tangent to a circle $\perp$ the radius through the point of contact] (1 Mark)
n$\therefore AC = CB$ [$\perp$ drawn from centre of a circle to a chord bisects the chord] (1 Mark)
n$OC = 4$cm and $OB = 5$ cm (1/2 Mark)
n$\therefore$ by Pythagoras theorem in $\Delta BOC$,
n$BC = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ cm (1/2 Mark)
n$\therefore AB = 2BC = 2 \times 3 = 6$ cm (1 Mark)