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(B) To find the equation of the pair of lines passing through the origin and the intersection points of the circle $x^2 + y^2 = 4$ and the line $2x +

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$5y^2 + 12xy = 0$

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(B) To find the equation of the pair of lines passing through the origin and the intersection points of the circle $x^2 + y^2 = 4$ and the line $2x + 3y - 4 = 0$,we homogenize the equation of the circle using the equation of the line.From the line equation,we have $2x + 3y = 4$,which implies $\frac{2x + 3y}{4} = 1$.Substituting this into the circle equation $x^2 + y^2 = 4(1)^2$:$x^2 + y^2 = 4 \left( \frac{2x + 3y}{4} \right)^2$$x^2 + y^2 = 4 \left( \frac{4x^2 + 9y^2 + 12xy}{16} \right)$$x^2 + y^2 = \frac{4x^2 + 9y^2 + 12xy}{4}$$4x^2 + 4y^2 = 4x^2 + 9y^2 + 12xy$$5y^2 + 12xy = 0$.

Find the equation of the pair of lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 4$ and the line $2x + 3y - 4 = 0$.

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