The equations of the tangents to the circle x | vedclass

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(C) The given circle is $x^2 + y^2 = 4$,which has its center at $(0, 0)$ and radius $r = 2$.Any line parallel to $x + 2y + 3 = 0$ can be written in th

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$x + 2y = \pm 2\sqrt{5}$

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(C) The given circle is $x^2 + y^2 = 4$,which has its center at $(0, 0)$ and radius $r = 2$.Any line parallel to $x + 2y + 3 = 0$ can be written in the form $x + 2y + \lambda = 0$.For this line to be a tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = 2$.The perpendicular distance $d$ from $(x_1, y_1)$ to $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get:$\frac{|1(0) + 2(0) + \lambda|}{\sqrt{1^2 + 2^2}} = 2$$\frac{|\lambda|}{\sqrt{5}} = 2$$|\lambda| = 2\sqrt{5}$$\lambda = \pm 2\sqrt{5}$Substituting $\lambda$ back into the equation of the line,we get $x + 2y \pm 2\sqrt{5} = 0$,or $x + 2y = \pm 2\sqrt{5}$.

The equations of the tangents to the circle $x^2 + y^2 = 4$,which are parallel to $x + 2y + 3 = 0$,are

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