Question

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle $x^2+y^2-x+3y=0$ on $L_1$ and $L_2$ are equal then which of the following equation can represent $L_1$?

Answer 1

$\because L_1$ passes through origin.

Let equation of $L_1$.$y=mx$

Intercepts made by $L_1$ and $L_2$ are equal

$\Rightarrow L_1$ and $L_2$ are at the same distance from centre of circle $x^2+y^2-x+3y=0$

$x^2+y^2-x+3y=0$

$x^2-x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+y^2+3y+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2=0$

${(x-\frac{1}{2})}^2+{(y+\frac{3}{2})}^2={\left(\frac{\sqrt{10}}{2}\right)}^2$

$\therefore$ Centre of circle $(\frac{1}{2},\frac{-3}{2})$

Distance of a point $(x_1,y_1)$ from line $Ax+By+C=0$ is given by

$d=\frac{|A{x}_1+B{y}_1+1|}{\sqrt{A^2+B^2}}$

$\therefore$ Distance of $(\frac{1}{2},\frac{-3}{2})$ from line $L_1$ and $L_2$ are

$d_1=\frac{|m\left(1/2\right)+3/2|}{\sqrt{m^2+1}}$

$d_2=\frac{|1/2+\left(3/2\right)-1|}{\sqrt{1+1}}$

$d_1=d_2$

$\Rightarrow \frac{m-3}{2\sqrt{m^2+1}}=\frac{3}{\sqrt{2}}$

Squaring both sides we get

$\frac{(m+3{)}^2}{4(m^2+1)}=\frac{4}{2}$

$(m-3{)}^2=8(m^2+1)$

$m^2+9+6m=8m^2+3$

$7m^2-6m-1=0$

$m=\frac{6\pm \sqrt{36+28}}{14}=\frac{+6\pm 8}{14}=+1,-\frac{1}{7}$

$\therefore L_1:-y+x=0$, $x+7y=0$.