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The equation to the tangents to the circle ${x^2} + {y^2} = 4$, which are parallel to $x + 2y + 3 = 0$, are
$x - 2y = 2$
$x + 2y = \pm \,2\sqrt 3 $
$x + 2y = \pm \,2\sqrt 5 $
$x - 2y = \pm \,2\sqrt 5 $
Solution
(c) Trick : Only $(b)$ and $(c)$ lines are parallel to $x + 2y + 3 = 0$
Also the line is a tangent to ${x^2} + {y^2} = 4$
Its distance from $(0, 0)$ should be $2$.
Therefore $c$ is the answer.
Alternative method :
Centre of ${x^2} + {y^2} = 4$ is $(0, 0).$
Tangents which are parallel to $x + 2y + 3 = 0$ is $x + 2y + \lambda = 0$…..$(i)$
Perpendicular distance from $(0,0)$ to $x + 2y + \lambda = 0$ should be equal to
radius of circle, (Clearly radius = $2$).
$\therefore $ $\frac{{0 + 2 \times \,0 + \lambda }}{{\sqrt {{1^2} + {2^2}} }} = \, \pm \,2$
$ \Rightarrow $ $\lambda \, = \pm \,2\sqrt 5 $
Put the value of $\lambda \,$ in $(i)$,
tangents of circle are $x + 2y = \, \pm \,2\sqrt 5 .$
