Question

Two sides of a triangle are along the coordination
axes and its hypotenuse passes through the point
of intersection of the lines px+qy+r=0px+qy+r=0

Answer 1

Any line through the intersection of given line is
$(px+qy+r)+\lambda (lx+my+n)=0$
It means the axes $y=0$ and $x=0$ at
$A(-\frac{r+\lambda n}{p+\lambda l},0),B(0-\frac{r+\lambda n}{q+\lambda m})$
Since the triangle is right angled, the circumcentre will be at the mid-point of hypotenuse $AB$.
If $(x,y)$ be the circmcentre then
$2x=-\frac{r+\lambda n}{p+\lambda l},2y=-\frac{r+\lambda n}{q+\lambda m}$
$\therefore \lambda (2xl+n)+(2px+r)=0$
and $\lambda (2ym+n)+(2qy+r)=0$
now eliminate $\lambda$ from the above two
$\frac{2xl+n}{2px+r}=\frac{2ym+n}{2qy+r}=(-\frac{1}{\lambda })$
Now cross multiply and cancel $2$ to get the locus as given.