What is the angle subtended by the common cho | vedclass

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(D) Given circles are $S_1: x^2 + y^2 - 4x - 4y = 0$ and $S_2: x^2 + y^2 - 16 = 0$.The equation of the common chord is $S_1 - S_2 = 0$,which gives $-4

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$\pi /2$

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(D) Given circles are $S_1: x^2 + y^2 - 4x - 4y = 0$ and $S_2: x^2 + y^2 - 16 = 0$.The equation of the common chord is $S_1 - S_2 = 0$,which gives $-4x - 4y + 16 = 0$,or $x + y - 4 = 0$.Thus,the equation of the common chord is $x + y = 4$,or $\frac{x + y}{4} = 1$.To find the pair of lines joining the origin to the intersection points of the circle $S_2 = 0$ and the chord,we homogenize the circle equation using the chord equation:$x^2 + y^2 - 16(1)^2 = 0$$x^2 + y^2 - 16\left(\frac{x + y}{4}\right)^2 = 0$$x^2 + y^2 - 16\left(\frac{x^2 + y^2 + 2xy}{16}\right) = 0$$x^2 + y^2 - (x^2 + y^2 + 2xy) = 0$$-2xy = 0$,which implies $xy = 0$.The lines are $x = 0$ (y-axis) and $y = 0$ (x-axis).The angle between the lines $x = 0$ and $y = 0$ is $\frac{\pi}{2}$.

What is the angle subtended by the common chord of the circles $x^2 + y^2 - 4x - 4y = 0$ and $x^2 + y^2 = 16$ at the origin?

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