Question

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

• A. $(x^2-y^2{)}^2=x^2+y^2$
• B. $(x^2+y^2{)}^2=(x^2+y^2)$
• C. $(x^2+y^2)=4x^2{y}^2$
• D. $(x^2-y^2{)}^2=(x^2+y^2{)}^2$

Answer 1

The correct option is A $(x^2-y^2{)}^2=x^2+y^2$

The triangle formed by the given three lines will be right-angled. (Since angle between $x+y=0$ and $x-y=0$ is ${90}^o$)

Hence, circumcenter will be the midpoint of the hypotenuse.

Solving $x-y=0$ and $lx-my=1$ we get

$x=y=\frac{1}{l-m}$

Similarly, solving $x+y=0$ and $lx-my=1$ we get

$x=-y=\frac{1}{l+m}$

Hence, coordinates of circumcentre, using midpoint formula, are

$\frac{l}{(l^2-m^2)},\frac{m}{(l^2-m^2)}$

Hence,
$h=\frac{l}{l^2-m^2}$ ......... (i)

$k=\frac{m}{l^2-m^2}$ ......... (ii)
Square and adding (i) and (ii), we get
$h^2+k^2=\frac{l^2+m^2}{(l^2-m^2{)}^2}=\frac{1}{(l^2-m^2{)}^2}$
(putting $l^2+m^2=1$)
$(h^2-k^2)=\frac{1}{(l^2-m^2)}$
Therefore, the locus is $x^2+y^2=(x^2-y^2{)}^2$.