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Q.
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
Conic Sections
Solution:
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 $x^{2} + y^{2} = 16\, is \,x+y = 4$ which meets the circle $x^{2 }+ y^{2} = 16$ at points $A\left(4, \,0\right)$ and $B\left(0,\, 4\right)$. Obviously $OA \bot OB$. 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.
