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 equations can represent $L_1$ ?

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

Answer 1

Answer: (B), (C)

Lines $L_1\equiv y-mx=0$
$L_2\equiv y+x-1=0$
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 : $8m^2+8=m^2+6m+9$
$7m^2-6m-1=0\Rightarrow m=-\frac{1}{7},1$
so lines are $y+\frac{1}{7}x=0\Rightarrow 7y+x=0$
and $y-1\cdot x=0\Rightarrow y-x=0$