Question

The equation of a chord of the circle $x^2+y^2-3x-4y-4=0$, which passes through the origin such that the origin divides it in the ratio $4:1$, is

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

Answer 1

The correct option is B $24x+7y=0$

Then point of intersections are given by

$x^2(1+m^2)-x(3+4m)-4=0$

$\therefore x_1+x_2=\frac{3+4m}{1+m^2}$ and $x_1{x}_2=\frac{-4}{1+m^2}$

Since $(0,0)$ divides the chord in the ratio $1:4,$

$\therefore x_2=-4x_1$

$-3x_1=\frac{3+4m}{1+m^2}$ and $4x_1^2=\frac{-4}{1+m^2}$

$\therefore 9+9m^2=9+16m^2+24m$

i.e., $m=0,-\frac{24}{7}$

Therefore, the lines are $y=0$ and $7y+24x=0$.