Question

Let $3x-y-3=0$ is a diameter of the circle $x^2+y^2-4x-6y+4=0$. If $L_1$ is the chord which is bisected by the given diameter line, then which of the following is/are true?

• A. When $L_1$ is the largest possible chord of the circle, its possible equation is $x+3y-11=0$
• B. When $L_1$ is the largest possible chord of the circle, its possible equation is $x-3y+8=0$
• C. When $L_1$ is at a distance of $1$ from the centre, its equation is $x+3y-11+\sqrt{10}=0$
• D. When $L_1$ is at a distance of $1$ from the centre, its equation is $x+3y-11-\sqrt{10}=0$

Answer 1

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

When $L_1$ is at a distance of $1$ from the centre, its equation is $x+3y-11-\sqrt{10}=0$

Given circle is $x^2+y^2-4x-6y+4=0$
$C=(2,3),r=3$
Diameter is $3x-y-3=0$
The largest chord is diameter and two diameter always bisect each other.
Checking the equations
$x+3y-11=0$ and $x-3y+8=0$
Diameter passes through the centre $(2,3)$, so
$2+9-11=02-9+8=1\ne 0$
Therefore $x+3y-11=0$ can be equation of diameter.

The chord which is bisected by the given diameter is perpendicular to it, so the equation of chord is
$x+3y+\lambda =0$
Distance from centre is $1$, so
$\frac{|11+\lambda |}{\sqrt{10}}=1\Rightarrow \lambda =-11\pm \sqrt{10}$
Hence, the required equation of chord are
$x+3y-11+\sqrt{10}=0x+3y-11-\sqrt{10}=0$