The equation to the tangents to the circle {x | vedclass

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Question

(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,

Vedclass · Accepted Answer

$x + 2y = \pm \,2\sqrt 5 $

Vedclass · Answer

(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 = $2$).

$	herefore $ $\frac{{0 + 2 	imes \,0 + \lambda }}{{\sqrt {{1^2} + {2^2}} }} = \, \pm \,2$

$ \Rightarrow $ $\lambda \, = \pm \,2\sqrt 5 $

Put the value of $\lambda \,$ in $(i)$,

tangents of circle are $x + 2y = \, \pm \,2\sqrt 5 .$

The equation to the tangents to the circle ${x^2} + {y^2} = 4$, which are parallel to $x + 2y + 3 = 0$, are

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