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 $x^2+y^2-x+3 y=0$ on $L_1$ and $L_2$ are equal, then which of the following equations can represent $L_1$ ?

• A. $x+y=0$
• B. $x-y=0$
• C. $x+7 y=0$
• D. $x-7 y=0$

Answer 1

Answer: (B), (C)

Lines $L_1 \equiv y-m x=0$
$L_2 \equiv y+x-1=0$
If two lines make equal intercept on same circle then perpendicular distance from centre $\left(\frac{1}{2},-\frac{3}{2}\right)$ to both the lines are same.
$ \Rightarrow \left|\frac{-3 / 2+1 / 2-1}{\sqrt{1^2+1^2}}\right|=\left|\frac{-3 / 2-m / 2}{\sqrt{1+m^2}}\right| $
$ \Rightarrow \frac{2}{\sqrt{2}}=\frac{\mid m+3|}{2 \sqrt{1+m^2}}$
$ \Rightarrow 2 \sqrt{2} \sqrt{1+m^2}=|m+3|$
Upon squaring : $8 m^2+8=m^2+6 m+9$
$7 m^2-6 m-1=0 \Rightarrow m=-\frac{1}{7}, 1$
so lines are $y+\frac{1}{7} x=0 \Rightarrow 7 y+x=0$
and $ y-1 \cdot x=0 \Rightarrow y-x=0$