Finding external ray of radius

Question

Since all radiI are equal in length I could find the radius part but the external part of radius I couldn't find it I will be thankful for help

Answer 1

use pythagorean theorem:

$MB=\sqrt{MN^2-NB^2}=\sqrt{(8+5)^2-5^2}=12$

$NC=\sqrt{MN^2-MC^2}=\sqrt{(8+5)^2-8^2}=\sqrt{105}$