Question

The equation of diameter which bisects the chord $3x+y+5=0$ of the circle $x^2+y^2=16$ is

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

Answer 1

The correct option is D $x-3y=0$

Given circle is $x^2+y^2=16$
Chord is $3x+y+5=0$
The diameter which bisects the chord is perpendicular to the chord and passes through centre, so
$x-3y+\lambda =0$
Putting the centre $(0,0)$
$0-0+\lambda =0\Rightarrow \lambda =0$
So, the equation of diameter is
$x-3y=0$