Problem: A(2, 0), B(0, 2) and C are the vertices of a triangle inscribed in the circle $\large x^2 + y^2 = 4 $ . The bisector of angle C belongs to either of two families of concurrent lines, whose points of concurrency are

(A) (√2 , √2 )

(B) (-√2 , -√2 )

(C) (√2 , -√2 )

(D) (-√2 , √2 )

Ans: (A), (B).

Sol: The points of concurrency lie on the intersection of perpendicular bisectors of AB with the circle x^{2} + y^{2} = 4.

Hence the points are (√2 , √2 ) or (-√2 , -√2 )