Eliminate $l, m$ between the equations $l x + m y = a$ , $m x - l y = b$ , $l^{2} + m^{2} = 1$ .

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Solution

$l x + m y = a$ ....... $(i)$
$m x - l y = b$ ....... $(i i)$
$l^{2} + m^{2} = 1$ ....... $(i i i)$
Let's square $(i)$ , $(i i)$ and add them
$(l x + m y)^{2} = a^{2}$
$(m x - l y)^{2} = b^{2}$
$(l^{2} x^{2} + m^{2} y^{2} + m^{2} x^{2} + l^{2} y^{2}) = (a^{2} + b^{2})$
$l^{2} (x^{2} + y^{2}) + m^{2} (x^{2} + y^{2}) = (a^{2} + b^{2})$
$(l^{2} + m^{2}) (x^{2} + y^{2}) = (a^{2} + b^{2})$
Let $l = cos t, m = sin t$
We know that ${cos}^{2} t + {sin}^{2} t = 1$
So, given constraint on $l, m$ is true.
Thus, the eliminated equation is $(x^{2} + y^{2}) = (a^{2} + b^{2})$

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