Consider the triangle formed by the lines x + | vedclass

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(A) The lines are $L_1: x+y=0$,$L_2: x-y=0$,and $L_3: lx+my=1$.Since $L_1$ and $L_2$ are perpendicular and pass through the origin $(0,0)$,the triangl

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$(x^2 - y^2)^2 = x^2 + y^2$

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(A) The lines are $L_1: x+y=0$,$L_2: x-y=0$,and $L_3: lx+my=1$.Since $L_1$ and $L_2$ are perpendicular and pass through the origin $(0,0)$,the triangle is a right-angled triangle with the right angle at the origin.The vertices are $O(0,0)$,$A = \left(\frac{1}{l+m}, \frac{1}{l+m}\right)$,and $B = \left(\frac{1}{l-m}, \frac{1}{m-l}\right)$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AB$.Let the circumcentre be $(h, k)$.$h = \frac{1}{2} \left(\frac{1}{l+m} + \frac{1}{l-m}\right) = \frac{1}{2} \left(\frac{l-m+l+m}{l^2-m^2}\right) = \frac{l}{l^2-m^2}$$k = \frac{1}{2} \left(\frac{1}{l+m} + \frac{1}{m-l}\right) = \frac{1}{2} \left(\frac{m-l+l+m}{m^2-l^2}\right) = \frac{m}{m^2-l^2} = -\frac{m}{l^2-m^2}$Now,$h^2 + k^2 = \frac{l^2+m^2}{(l^2-m^2)^2} = \frac{1}{(l^2-m^2)^2}$ (since $l^2+m^2=1$).Also,$h^2 - k^2 = \frac{l^2-m^2}{(l^2-m^2)^2} = \frac{1}{l^2-m^2}$.Thus,$h^2 + k^2 = (h^2 - k^2)^2$.The locus is $x^2 + y^2 = (x^2 - y^2)^2$.

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

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