Question

2. 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.
(CBSE 2013]​

Answer 1

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✤ Required Answer:

⚡GiveN:

• There are two concentric circles.
• Chord of larger circle touching smaller circle = 8 cm
• Radius of larger circle = 5 cm

⚡To FinD:

• Radius of the smaller circle.....$?$

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✤ How to solve?

We know that,

In this circle, the chord of larger circle is a tangent to the smaller circle and radius is always perpendicular to the point of tangency(The point where the tangent touches the circle). And, also the perpendicular bisects the chord.

❍ So, let's solve this question....

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✤ Solution:

According to above,

• OB perpendicular to AB
• AB = BP (They are bisected)

[ So, Triangle OAB is a right angled triangle]

We have,

• OA = 5 cm (Radius of larger circle)
• AB = 4 cm (Bisected)

In Right angled triangle OAB,

But using pythagoras theoram,

➝ p² + b² = h²

➝ OB² + AB² = OA²

➝ OB² + 4² = 5²

➝ OB = √5² - 4² cm

➝ OB = √25 - 16 cm

➝ OB = √9 cm

➝ OB = 3 cm

❒ Radius of the larger circle = 3 cm

Hence, solved !!

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Answer 2

$\huge\bf\therefore \color{red} Given : -$


AB = 8cm
OA = 5cm



⇢ 'O' is common centre and as ; AB is chord and 'C' is touch point of inner circle by chord AB;


⇢ 'CO ⊥ AB'


∴ By Theorem

'C' bisect The AB ;

∴ AC = CB { 'C' is mid pointed of AB }


$\sf \: ⇢ AC = \frac{1}{2} \: AB \\ \\ \sf \: AC = \frac{1}{2} \big(8 \big) = 4cm \\ \\ \therefore \sf In Right \ \: \triangle OAB \\ \\ ⇢ \sf OA^2 = OC^2 + AC^2$