Find the equation of tangents to circle $x^{2} + y^{2} = 4$ which make an angle of $60^{o}$ with the $x$ -axis.

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Solution

$t a n g e n t m a k e s 60^{0} w i t h w i l l X - a x i s$
$∴ s l o p e o f t a n g e n t = t a n 60^{0} = \sqrt{3} n o w x^{2} + y^{2} = 4 . . . . . . . (i)$
$2 x + 2 y (\frac{d y}{d x}) = 0$
$o r (\frac{d y}{d x}) = (\frac{- x}{y}) o r \sqrt{3} y = (\frac{- x}{y})$
$∴ x = - \sqrt{3} y . . . . . . (i i)$
$u s e e q^{n} (i i) i n e q^{n} (i), w e g e t p o i n t o f t a n g e n t a t c i r c l e$
$(- \sqrt{3} y)^{2} + y^{2} = 4$
$o r 3 y^{2} + y^{2} = 4 ∴ y^{2} = 1 o r y = \pm 1$
$t h e n x = \mp \sqrt{3}$
$e q^{n} o f t a n g e n t (\frac{y - 1}{x + \sqrt{3}}) = \sqrt{3} o r y - 1 = \sqrt{3} x + 3 o r y = \sqrt{3} x + 4$
$(\frac{y + 1}{x - \sqrt{3}}) = \sqrt{3} o r y + 1 = \sqrt{3} x - 3 o r y = \sqrt{3} x - 4$

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