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

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Aicharya · Accepted Answer

Then, equation of common chord in given by $ \begin{aligned} & S_1-S_2 & =0 \\ & \Rightarrow\left(x^2+y^2-4 x+8 y+5\right) & \\ & & \\ \Rightarrow & -\left(x^2+y^2+2 x+4 y-3\right) & =0 \\ \Rightarrow & -6 x+4 y+8 & =0 \\ \Rightarrow & -3 x+2 y+4 & =0 \\ & 3 x-2 y & =4 \end{aligned} $

Now, the required equation of pair of lines is given by homogenization of Eqs. (i) and (iii) or Eqs. (ii) and (iii) On homogenization of Eqs. (i) and (iii), we get $ \begin{aligned} x^2+y^2-4 x\left(\frac{x}{4 / 3}+\frac{y}{-2}\right)+8 y\left(\frac{x}{4 / 3}\right. & \left.+\frac{y}{-2}\right) \\ +5\left(\frac{x}{4 / 3}+\frac{y}{-2}\right)^2 & =0 \end{aligned} $ $\begin{aligned} & \Rightarrow x^2+y^2-4 x\left(\frac{3}{4} x-\frac{y}{2}\right)+8 y\left(\frac{3}{4} x+\frac{y}{-2}\right) \\ & +5\left(\frac{3 x}{4}-\frac{y}{2}\right)^2=0 \\ & \Rightarrow x^2+y^2-3 x^2+2 x y+6 x y-4 y^2 \\ & +5\left(\frac{9}{16} x^2+\frac{y^2}{4}-\frac{3 x y}{4}\right)=0 \\ & \Rightarrow \quad-2 x^2-3 y^2+8 x y+\frac{45}{16} x^2+\frac{5}{4} y^2-\frac{15 x y}{4}=0 \\ & \Rightarrow \quad-32 x^2-48 y^2+128 x y+45 x^2+20 y^2-60 x y=0 \\ & \Rightarrow \quad 13 x^2-28 y^2+68 x y=0 \\ & \end{aligned}$

The equation of the pair of lines joining the origin to the points of intersection of two circles $x^2+y^2-4…

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