Find the equation of the circle whose centre is the point of intersection of the lines $2x-3y+4=0$ and $3x+4y-5=0$ and passes through the origin.

The point of intersection of the lines $2x-3y+4=0$ and $3x+4y-5=0$ is $\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$

Equation of circle whose centre is $\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$ and passing through origin is

$x^2+y^2+\displaystyle\frac{2x}{17}-\displaystyle\frac{44y}{17}=0$

$\Rightarrow 17(x^2+y^2)+2x-44y=0$

Hence, option A.