Two circles of radii $10cm$ and $8cm$ intersect each other, and the length of the common chord is $12cm.$ The distance between their centers will be

The correct option is B: $8+2\sqrt{7}cm$

$OA=10cm,CA=8cm$ and $AB=12cm$

$AD=\frac{1}{2}AB=6cm$ ($\perp$ from the center of the circle bisects the chord)

In right triangle OAD, we have

$O{A}^2=O{D}^2+A{D}^2$

$\Rightarrow {10}^2=O{D}^2+6^2$

$\Rightarrow O{D}^2=\sqrt{100-36}=8cm$

Again, in right triangle ADC, we have

$A{C}^2=A{D}^2+D{C}^2$

$\Rightarrow 8^2=6^2+D{C}^2$

$\Rightarrow DC=\sqrt{64-36}=\sqrt{28}=2\sqrt{7}cm$

$\therefore OC=OD+DC=8+2\sqrt{7}$

Hence, the distance between the centers is $(8+2\sqrt{7})cm$