# The common chord of $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ subtends at the origin an angle equal to

Problem: The common chord of $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ subtends at the origin an angle equal to

(A) π/6

(B) π/4

(C) π/3

(D) π/2

Sol. The equation of the common chord of the circles $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ is

x + y = 4 which meets the circle x2 + y2 = 16 at points A(4,0) and B(0,4).

Obviously OA ⊥ OB.

Hence the common chord AB makes a right angle at the centre of the circle

x2 + y2 = 16

Hence (D) is the correct answer.