If $4 l^{2} - 5 m^{2} + 6 l + 1 = 0$ , prove that the line $l x + m y + 1 = 0$ touches a fixed circle.

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Solution

Let the circle be $(x - a)^{2} + (y - b)^{2} = r^{2}$ .
Condition of tangency gives
$\frac{(a l + b m + 1)}{\sqrt{l^{2} + m^{2})}} = r$
or $(a l + b m)^{2} + 1 + 2 (a l + b m) = r^{2} (l^{2} + m^{2})$
or $l^{2} (a^{2} - r^{2}) + m^{2} (b^{2} - r^{2}) + 2 a b l m + 2 a l + 2 b m + 1 = 0 . . . . . (1)$
Compare with $4 l^{2} - 5 m^{2} + 6 l + 1 = 0$
$∴ a^{2} - r^{2} = 4, b^{2} - r^{2} = - 5, a b = 0$ ,
$2 a = 6, 2 b = 0$
$∴ a = 3, b = 0, ∴ r^{2} = 5$
Hence the circle is $(x - 3)^{2} + y^{2} = 5$
or $x^{2} + y^{2} - 6 x + 4 = 0$ .

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