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

Let y = mx be a chord.
Then the points of intersections are given by
$\begin{array}{l}{\mathrm{x}}^2\left(1+{\mathrm{m}}^2\right)-\mathrm{x}(3+4\mathrm{m})-4=0\\ \therefore {\mathrm{x}}_1+{\mathrm{x}}_2=\frac{3+4\mathrm{m}}{1+{\mathrm{m}}^2}\text{ and }{\mathrm{x}}_1{\mathrm{x}}_2=\frac{-4}{1+{\mathrm{m}}^2}\end{array}$
Since (0, 0) divides chord in the ratio 1 : 4, we have
$\begin{array}{l}{\mathrm{x}}_2=-4{\mathrm{x}}_1\\ -3{\mathrm{x}}_1=\frac{3+4\mathrm{m}}{1+{\mathrm{m}}^2}\text{ and }4{\mathrm{x}}_1^2=-\frac{-4}{1+{\mathrm{m}}^2}\\ 9+9{\mathrm{m}}^2=9+16{\mathrm{m}}^2+24\mathrm{m}\\ \text{ i.e., }\mathrm{m}=0,-\frac{24}{7}\end{array}$
Therefore, the lines are y= 0 and 7y + 24 x = 0.