Question

Two equal chords of the circle $x^2+y^2-2x+4y=0$, passing through the origin are perpendicular to each other; their equations are

• A. $x-2y=0$
• B. $2x+y=0$
• C. $x+3y=0$
• D. $3x-y=0$

Answer 1

Answer: (C), (D)

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

Let the chords be $OA$ and $OB$, then $OA=OB$ and $OA$ is perpendicular to $AB\Rightarrow AB$ is a diameter perpendicular to the line joining the origin $O$ and the center $(1,-2)$.

So its equation is $x-2y-5=0$.

Equation of $OA$ and $OB$ are ontained by making the equation of the circle homogeneous with the help of the equation of $AB$

$x^2+y^2-2(x-2y)\left(\frac{x-2y}{5}\right)=0\Rightarrow 3x^2+8xy-3y^2=0\Rightarrow (3x-y)(x+3y)=0$