Question

Let $C:x^2+y^2-3x-4y-4=0.$ A chord of $C$ passes through the origin such that the origin divides it in the ratio $4:1$. Then the equation(s) of chord is/are

• A. $x=0$
• B. $y=0$
• C. $7x+24y=0$
• D. $24x+7y=0$

Answer 1

Answer: (B), (D)

The correct options are
B $y=0$
D $24x+7y=0$

Let equation of chord is $y=mx$
$\text{Then}{x}^2+(mx{)}^2-3x-4mx-4=0\Rightarrow x^2(1+m^2)-x(3+4m)-4=0$

Let $\alpha ,\beta$ be the $x-$coordinate of the intersection point of the chord and the circle.
$\text{Then}\alpha +\beta =\frac{3+4m}{1+m^2}\text{and}\alpha \beta =\frac{-4}{1+m^2}$

Since the origin divides the chord in the ratio of $4:1$
$4\times \alpha +1\times \beta =0$
$\Rightarrow \beta =-4\alpha$
$\Rightarrow -3\alpha =\frac{3+4m}{1+m^2},-4{\alpha }^2=\frac{-4}{1+m^2}\Rightarrow m=0,\frac{-24}{7}$
Hence, the required lines are $y=0$ and $24x+7y=0$