Question : There are two circles which touch each other externally. The radius of the first circle with centre O is 17 cm and the radius of the second circle with centre A is 7 cm. BC is a direct common tangent to these two circles, where B and C are points on the circles with centres O and A, respectively. The length of BC is:
Option 1: $2 \sqrt{118} $ cm
Option 2: $2 \sqrt{119} $ cm
Option 3: $2 \sqrt{113} $ cm
Option 4: $2 \sqrt{117}$ cm
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Correct Answer: $2 \sqrt{119} $ cm
Solution : Since AB $\perp$ BC and AQ $\perp$ OC, AQBC is a rectangle with sides AQ = BC and AB = QC. ⇒ AP = AB = QC = 7 cm and OC = OP = 17 cm Also, AO = AP + PO = 7 +17 = 24 cm and OQ = CO – CQ = 17 – 7 = 10 cm In $\triangle$ AOQ, AO 2 = AQ 2 + OQ 2 ⇒ 24 2 = AQ 2 + 10 2 ⇒ 576 = AQ 2 + 100 ⇒ AQ = $\sqrt{476}$ = $2\sqrt{119}$ ⇒ BC = AQ = $2\sqrt{119}$ Hence, the correct answer is $2\sqrt{119}$ cm.
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