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(D) To find the lines joining the origin to the points of intersection,we homogenize the equation of the circle using the line equation.Given $x + y =

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All of the above

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(D) To find the lines joining the origin to the points of intersection,we homogenize the equation of the circle using the line equation.Given $x + y = 2$,we have $\frac{x + y}{2} = 1$.Substituting this into the circle equation $x^2 + y^2 = 3(1)^2$:$x^2 + y^2 = 3(\frac{x + y}{2})^2$$x^2 + y^2 = \frac{3}{4}(x^2 + 2xy + y^2)$$4x^2 + 4y^2 = 3x^2 + 6xy + 3y^2$$x^2 - 6xy + y^2 = 0$Dividing by $y^2$,we get $(\frac{x}{y})^2 - 6(\frac{x}{y}) + 1 = 0$.Using the quadratic formula for $\frac{x}{y}$:$\frac{x}{y} = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm \sqrt{8} = 3 \pm 2\sqrt{2}$.Thus,$x = (3 + 2\sqrt{2})y$ and $x = (3 - 2\sqrt{2})y$.These can be written as $x - (3 + 2\sqrt{2})y = 0$ and $x - (3 - 2\sqrt{2})y = 0$.Also,$x = (3 + 2\sqrt{2})y \Rightarrow y = (3 - 2\sqrt{2})x$ (since $(3+2\sqrt{2})(3-2\sqrt{2}) = 1$),which leads to $y - (3 - 2\sqrt{2})x = 0$ and $y - (3 + 2\sqrt{2})x = 0$.Since all forms are equivalent representations of the pair of lines,the correct option is $D$.

The lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 3$ and the line $x + y = 2$ are:

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