Question

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. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

The centre of two circles are $C_1(2,2)$ and $C_2(0,0)$. The radii of two circles are $r_1=2\sqrt{2}$ and $r_2=4$

The eq. of the common chord of the circles $x^2+y^2-4x-4y=0$ and $x^2+y^2=16isx+y=4$ which meets the circle $x^2+y^2=16$ at points $A\left(4,0\right)$ and $B\left(0,4\right)$. Obviously $OA\perp OB$. Hence, the common chord AB makes a right angle at the centre of the $x^2+y^2=16$. Where, O is the origin and the centre $C_2$ of the second circle.