Question : Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Then the distance between their centres is:
Option 1: 10 cm
Option 2: 8 cm
Option 3: 13.3 cm
Option 4: 15 cm
Correct Answer: 13.3 cm
Solution :
Let O and P be the centre of two circles with radii 10 cm and 8 cm.
OA = 10 cm, PA = 8 cm, and AB = 12 cm
Then AC = $\frac{1}{2}$ × 12 = 6 cm
In $\triangle OCA$,
$OA^2=OC^2+AC^2$
⇒ $10^2=OC^2+6^2$
⇒ $100=OC^2+36$
⇒ $OC^2 = 64$
$\therefore OC=8$
Now, In $\triangle ACP$,
$AP^2=AC^2+CP^2$
⇒ $8^2=6^2+CP^2$
⇒ $64=36+CP^2$
⇒ $CP^2=28$
$\therefore CP=2\sqrt7 =2×2.65= 5.3$ cm [$\because \sqrt7=2.65$]
So, total distance $=OP=OC+CP = 8 + 5.3 = 13.3$ cm
Hence, the correct answer is 13.3 cm.
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