1.

Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal toA. 4 cmB. 6 cmC. 8 cm D. 10 cm

Answer»

Answer is C. 8 cm

Given: 

OA = 5 cm 

OQ = 3 cm 

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By property 1, ∆OAQ is right-angled at ∠OQA (i.e., ∠OQA = 90°). 

By Pythagoras theorem in ∆OAQ, 

OA2 = QA2 + OQ2

⇒ QA2 = OA2 – OQ2 

⇒ QA2 = 52 – 32 

⇒ QA2 = 252 – 92 

⇒ QA2 = 16

 ⇒ QA = √16 

⇒ QA = 4 cm 

By property 2,

BQ = BP (tangent from B) 

And, 

AQ = BQ = 4 cm [∵ Q is midpoint of AB] 

PB = PC = 4 cm [∵ P is midpoint of BC] 

Now, 

BC = BP + PC 

⇒ BC = 4 cm + 4 cm 

⇒ BC = 8 cm 

Hence, BC = 8 cm


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