The common chord of the circles x^{2}+y^{2}-4 | vedclass

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(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.Simplifying the second equation, we get $x^{2}+y^{2}=16$.The e

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$\frac{\pi}{2}$

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(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.Simplifying the second equation, we get $x^{2}+y^{2}=16$.The equation of the common chord is obtained by subtracting the two equations: $(x^{2}+y^{2}-4x-4y) - (x^{2}+y^{2}-16) = 0$.This simplifies to $-4x-4y+16=0$, or $x+y=4$.Let the origin be $O(0,0)$. The circle $x^{2}+y^{2}-4x-4y=0$ passes through the origin. The common chord $x+y=4$ intersects the circle at points $A$ and $B$.Since the circle $x^{2}+y^{2}-4x-4y=0$ can be written as $(x-2)^{2}+(y-2)^{2}=8$, its center is $C(2,2)$ and radius is $r = \sqrt{8} = 2\sqrt{2}$.The distance from the center $C(2,2)$ to the line $x+y-4=0$ is $d = \frac{|2+2-4|}{\sqrt{1^{2}+1^{2}}} = 0$.Since the distance is $0$, the common chord is a diameter of the first circle.Any diameter subtends a right angle at any point on the circumference. Since the origin $(0,0)$ lies on the circle, the angle subtended by the diameter at the origin is $\frac{\pi}{2}$.

The common chord of the circles $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$ subtends at the origin an angle equal to

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