Diameter CD and chord AB are mutually perpendicular and intersect at point E, CE = 4 AB + CE = CD

Diameter CD and chord AB are mutually perpendicular and intersect at point E, CE = 4 AB + CE = CD find the radius of the circle.

Chord AB intersects at right angles with the diameter CD of the circle, therefore, AE = BE.

By condition, AB + CE = CD.

CD = 4 + AB = 4 + 2 * AE.

Since the product of the segments formed at the intersection of two chords are equal, then

AE * BE = CE * DE.

AE ^ 2 = CE * (CD – CE).

AE ^ 2 = 4 * (4 + 2 * AE – 4).

AE ^ 2 = 8 * AE.

AE = 8 cm.

Then CD = 4 + 2 * AE = 4 + 2 * 8 = 20 cm.

СD is the diameter of the circle, then R = СD / 2 = 20/2 = 10 cm.

Answer: The radius of the circle is 10 cm.