​Let ABC be a triangle with\angle BAC = 90. A circle is tangent to the sides AB and AC at X and Y respectively, such that the points

Let the center of a circle be O and the midpoint of BC be M

Angle B = 45º

AM = sin( 45º) * 6

XY = 2/3 * AM            (Segment XY is a side of a square inscribed in a circle.)

Area of an inscribed square    A_s = XY²

r = sin(45º) * XY

Area of a circle     A_c = r²pi

Area of a shaded segment    A = (A_c - A_s) / 4 = pi - 2   

Find the area of the portion of the circle that lies outside the triangle.