Exams › SSC CGL (Prelims) › General
Correct answer: 17 cm
The radius to the point of tangency is perpendicular to the tangent, so the center, tangency point, and external point form a right triangle. With tangent length 15 cm and radius 8 cm, the distance from the external point to the center is $\sqrt{15^2+8^2}=\sqrt{289}=17$ cm.