Two circles intersect each other such that each passes through the centre of other. Radius of these circles are equal. If the length of common chord is $14\sqrt{14}$ cm, then what will be the sum of circumference of these circles ?
Answer & explanation
Correct answer: option 3
Let the radius be r cm.
According to the diagram,
XP = PY = 7\(\sqrt {3 }\)
OO' = r
Again O'X = OX = r (radius)
\(\angle\)XPO = \({90}^\circ\)
O'P = PO = \(\frac{r}{2}\)
\( { XO'}^{2 } \) = \( { O'P}^{2 } \) + \( { AP}^{2 } \)
\( { r}^{2 } \) = \( { r/2}^{2 } \) + \( { 7√3}^{2 } \)
⇒ \( { r}^{2 } \) = \( { r/4}^{2 } \) + 147
⇒ \( { 3r/4}^{2 } \) = 147
⇒ r = 7 x 2 = 14 cm,
Circumference of the two circles = 2 x 2 x \(\frac{22}{7}\) x 14
⇒ 176 cm.