A triangle is formed by the lines $x+y=0$ and $lx+my=1$ . If l and m vary subject to the condition$l^2+m^2=1$ , then the locus of its circum centre is

Coordinates of circum centre are $l/\left(l^2-m^2\right),m/\left(m^2-l^2\right)$ .

Hence

$h=\frac{l}{l^2-m^2}\mathrm{...............}\left(1\right)$

$k=-\frac{m}{l^2-m^2}\mathrm{...............}\left(2\right)$

Squaring and adding (1) and (2) we get

$h^2+k^2=\frac{l^2+m^2}{{\left(l^2-m^2\right)}^2}=\frac{1}{{\left(l^2-m^2\right)}^2}$ (putting $l^2+m^2=1$ )

Hence, $\frac{1}{{\left(l^2-m^2\right)}^2}={\left(h^2-k^2\right)}^2$

Therefore, the locus is ${\left(x^2-y^2\right)}^2=x^2+y^2$