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
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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