In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.

Let us consider circles C_{1} and C_{2} with common center as O. Let AB be a tangent to circle C_{1} at point P and chord in circle C_{2}. Join OB

In △OPB

OP ⏊ AB

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem,

[i.e. (Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}]

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

Now, 2PB = AB

[As we have proved above that OP ⏊ AB and Perpendicular drawn from center to a chord bisects the chord]

2PB = 8 cm

PB = 4 cm

(OB)^{2} = (5)^{2} + (4)^{2}

[As OP = 5 cm, radius of inner circle]

(OB)^{2} = 41

⇒ OB = √41 cm

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