Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Given: Two concentric circles (say C_{1} and C_{2}) with common center as O and radius r_{1} = 6.5 cm and r_{2} = 2.5 cm respectively.

To Find: Length of the chord of the larger circle which touches the circle C_{2}. i.e. Length of AB.

As AB is tangent to circle C_{2} and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2} ]

We have,

(OP)^{2} + (PB)^{2} = (OB)^{2}

r_{2}^{2} + (PB)^{2} = r_{1}^{2}

(2.5)^{2} + (PB)^{2}= (6.5)^{2}

6.25 + (PB)^{2} = 42.25

(PB)^{2} = 36

PB = 6 cm

Now, AP = PB ,

[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]

So,

AB = AP + PB = PB + PB

= 2PB = 2(6)

= 12 cm

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