The portion of the line $l x + m y = 1$ intercepted by the circle $x^{2} + y^{2} = a^{2}$ subtends an angle of $45^{o}$ at the centre of the circle. Prove that $4 [a^{2} (l^{2} + m^{2}) - 1] = [a^{2} (l^{2} + m^{2}) - 2]^{2}$ .

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Solution

The centre of the circle is origin. The equations of the lines joining origin to the points of intersection of the line and circle are
$x^{2} + y^{2} = a^{2} (l m + m y)^{2}$
or $x^{2} (a^{2} l^{2} - 1) + 2 l m a^{2} x y + y^{2} (a^{2} m^{2} - 1) = 0$
$A x^{2} + 2 H x y + B y^{2} = 0$
$tan 45^{o} = 1 = \frac{2 \sqrt{H^{2} - A B}}{A + B}$
$(A + B)^{2} = 4 [H^{2} - A B]$
$[a^{2} (l^{2} + m^{2}) - 2]^{2} = 4 [a^{2} (l^{2} + m^{2}) - 1]$ .

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