The equation of the pair of lines joining the origin to the points of intersection of two circles $x^{2} + y^{2} - 4 x + 8 y + 5 = 0 a n d x^{2} + y^{2} + 2 x + 4 y - 3 = 0 i s$

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

$x y - 28 y^{2} = 0$

$(x + 4) (x - 5) = 0$

$13 x^{2} + 6 x y - 28 y^{2} = 0$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n x^{2} + y^{2} - 4 x + 8 y + 5 = 0 - - - (1) \\ a n d x^{2} + y^{2} + 2 x + 4 y - 3 = 0 - - - (2) \\ E q u a t i o n o f c o m m o n c h o r d i s (1) - (2) \\ \Rightarrow - 6 x + 4 y + 6 = 0 \\ \Rightarrow 3 x - 2 y - 4 = 0 \\ \Rightarrow \frac{3 x - 2 y}{4} = 1 - - (3) \\ H o m o g e n i z i n g (2) w i t h (3) \\ x^{2} + y^{2} + 2 x (\frac{3 x - 2 y}{4}) + 4 y (\frac{3 x - 2 y}{4}) - 3 {(\frac{3 x - 2 y}{4})}^{2} = 0 \\ \Rightarrow 16 x^{2} + 16 y^{2} + 24 x^{2} - 16 x y + 48 x y - 32 y^{2} - 27 x^{2} - 12 y^{2} + 36 x y = 0 \\ \Rightarrow 13 x^{2} + 68 x y - 28 y^{2} = 0 \end{array}$