Question

Which of the following lines subtends chords of equal lengths on intersecting with the circle, $x^2+y^2-2x+4y=0$?

• A. $3x-y=0$
• B. $x+3y=0$
• C. $x+3y+10=0$
• D. $3x-y-10=0$

Answer 1

Answer: (A), (B), (C), (D)

$3x-y-10=0$

Given circle is $x^2+y^2-2x+4y=0$
Finding the point of intersection of the lines and circle, we get


For $3x-y=0$,
$x^2+(3x{)}^2-2x+4(3x)=0\Rightarrow 10x^2+10x=0\Rightarrow 10x(x+1)=0\Rightarrow x=-1,0\Rightarrow y=-3,0$
Distance between $(-1,-3)$ and $(0,0)$
$=\sqrt{1+9}=\sqrt{10}$

For $x+3y=0$,
$(-3y{)}^2+y^2-2(-3y)+4y=0\Rightarrow 10y(y+1)=0\Rightarrow y=-1,0\Rightarrow x=3,0$
Distance between $(3,-1)$ and $(0,0)$
$=\sqrt{10}$

Now, the centre and radius of the circle are
$C=(1,-2)r=\sqrt{5}$
We know that
$AM=MB\Rightarrow AB=2MB\Rightarrow AB=2\sqrt{O{B}^2-O{M}^2}\Rightarrow AB=2\sqrt{r^2-O{M}^2}\Rightarrow AB=2\sqrt{5-O{M}^2}$
$OM$ is the perpendicular distance from centre to the line

For $x+3y+10=0$
$OM=\left|\frac{1-6+10}{\sqrt{1^2+3^2}}\right|=\frac{5}{\sqrt{10}}AB=2\sqrt{5-\frac{25}{10}}\Rightarrow AB=\sqrt{10}$

For $3x-y-10=0$
$OM=\left|\frac{3+2-10}{3^2+1^2}\right|=\frac{5}{\sqrt{10}}\Rightarrow AB=\sqrt{10}$