From point B to the circle with center O, tangents BC and BD are drawn (D and C are tangency points).

From point B to the circle with center O, tangents BC and BD are drawn (D and C are tangency points). The segments BO and CD intersect at point N. Find the radius of the circle if CD = 16, BN = 4 4/15

The radii OD and OC are perpendicular to the tangents BD and BC. ОD = OC = R, then the right-angled triangles BOD and BOC are equal in leg and hypotenuse, then the angle OBD = OBC.

Since the angle ОВD = ОВC, then ВН is the height, bisector and median of the triangle ВСD.

Then CH = BC / 2 = 16/2 = 8 cm.

In a right-angled triangle BCH, CH there is a height drawn to the hypotenuse, then CH ^ 2 = BH * OH.

OH = CH ^ 2 / BH = 64 / (64/15) = 15 cm.

In a right-angled triangle SON, according to the Pythagorean theorem, CO ^ 2 = CH ^ 2 + OH ^ 2 = 64 + 225 = 289.

CO = R = 17 cm.

Answer: The radius of the circle is 17 cm.