A triangle is formed by the lines `x+y=0,x-y=0,` and `l x+m y=1.` If `la n dm` vary subject to the condition `l^2+m^2=1,` then the locus of its circumcenter is `(x^2-y^2)^2=x^2+y^2` `(x^2+y^2)^2=(x^2-y^2)` `(x^2+y^2)^2=4x^2y^2` `(x^2-y^2)^2=(x^2+y^2)^2`

Options

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

A

Explanation

The coordinates of circumcenter are `(l//(l^(2)-m^(2)), m//(m^(2)-l^(2))).`
Hence,
`h= (l)/(l^(2)-m^(2)) " " (1)`
`k= -(m)/(l^(2)-m^(2)) " " (2)`

Squaring and adding (1) and (2), we get
`h^(2)+k^(2) = (l^(2) + m^(2))/((l^(2)-m^(2))^(2)) = (1)/((l^(2)-m^(2))^(2)) " " ("Putting " l^(2) + m^(2)=1)`
`"Also, "h^(2)-k^(2) = (1)/(l^(2)-m^(2))`
Therefore, the locus is `x^(2) + y^(2) = (x^(2)-y^(2))^(2)`.