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(D) Let the two circles be $S_1: x^2+y^2-4x+8y+5=0$ and $S_2: x^2+y^2+2x+4y-3=0$. The common chord is given by $S_1-S_2=0$. $(x^2+y^2-4x+8y+5) - (x^2+

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$13x^2+68xy-28y^2=0$

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(D) Let the two circles be $S_1: x^2+y^2-4x+8y+5=0$ and $S_2: x^2+y^2+2x+4y-3=0$. The common chord is given by $S_1-S_2=0$. $(x^2+y^2-4x+8y+5) - (x^2+y^2+2x+4y-3) = 0$ $-6x+4y+8=0 \Rightarrow 3x-2y-4=0$. Thus,the equation of the line is $3x-2y=4$,or $\frac{3x-2y}{4}=1$. To find the pair of lines joining the origin to the intersection points,we homogenize the equation of $S_2$ using the line equation: $x^2+y^2+(2x+4y)(1) - 3(1)^2 = 0$ $x^2+y^2+(2x+4y)(\frac{3x-2y}{4}) - 3(\frac{3x-2y}{4})^2 = 0$ Multiplying by $16$ to clear the denominator: $16(x^2+y^2) + 4(2x+4y)(3x-2y) - 3(9x^2+4y^2-12xy) = 0$ $16x^2+16y^2 + 4(6x^2-4xy+12xy-8y^2) - 27x^2-12y^2+36xy = 0$ $16x^2+16y^2 + 24x^2+32xy-32y^2 - 27x^2-12y^2+36xy = 0$ $(16+24-27)x^2 + (32+36)xy + (16-32-12)y^2 = 0$ $13x^2+68xy-28y^2=0$.

The equation of the pair of lines joining the origin to the points of intersection of two circles $x^2+y^2-4x+8y+5=0$ and $x^2+y^2+2x+4y-3=0$ is

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