Question

Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

Answer 1

Given: OA = 10 cm, O'A = 8 cm and AB = 12 cm
$AD\mathit{=}\left(\frac{\mathrm{AB}}{2}\right)=\left(\frac{12}{2}\right)=6\mathrm{cm}$
Now, in right angled ΔADO, we have:
OA² = AD² + OD²
⇒ OD² = OA² - AD²
= 10² - 6²
= 100 - 36 = 64
∴ OD = 8 cm

Similarly, in right angled ΔADO', we have:
O'A² = AD² + O'D²
⇒ O'D² = O'A² - AD²
= 8² - 6²
= 64 - 36
= 28
⇒ $O\mathit{\text{'}}D=\sqrt{28}=2\sqrt{7}$ cm
Thus, OO' = (OD + O'D)
= $\left(8+2\sqrt{7}\right)\mathrm{cm}$
Hence, the distance between their centres is $\left(8+2\sqrt{7}\right)\mathrm{cm}$.