Points A, B, and C lie on a circle centered on point O. If line ZY is

The area of ∆ ZOY can be found in many ways. Here: ∆ ZOY consists of two congruent right isosceles triangles.
Their summed bases equal the base of ∆ ZOY, and their shared side, OB, with a known length, is the height of ∆ ZOY

For large ∆ ZOY, the third angle, at vertex Z, must = 45°
∆ ZOY already has one 90 and one 45-degree angle (given)
Angle at vertex Z: (180 - 135) = 45

Large ∆ ZOY consists of two congruent right isosceles triangles, ∆ BOY and ∆ BOZ
-- Tangent line YZ is perpendicular to radius OB, creating two right angles, one in each small triangle
-- Each small triangle has a vertex outside the circle, at Z (derived) and at Y (given), that = 45°
-- Their vertices at the center of the circle, O, both = 45° (for both small triangles, 90 + 45 + third vertex at O = 180, third vertex = 45)

Sides opposite equal angles are equal
Each small triangle has two equal sides, and both pairs (OB/BZ and OB/BY) lie opposite equal 45° angles.
Hence the sides are equal: Radius / side OB (shared) = side BZ = side BY

Length of those sides = \(4\)
Circle area = \(πr^2 = 16π , r^2 = 16 , r = 4 =\) OB, BZ, and BY

Base of ∆ ZOY = \((4 + 4) = 8\)
Area of ∆ ZOY = \(\frac{b*h}{2} = \frac{8*4}{2} = 16\)

Answer C