The equations of the tangents to the circle x | vedclass

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(A) The equation of any line parallel to $\sqrt{3}x + y + 3 = 0$ is of the form $\sqrt{3}x + y + k = 0$. Since this line is a tangent to the circle $x

Vedclass · Accepted Answer

$\sqrt{3}x + y \pm 2a = 0$

Vedclass · Answer

(A) The equation of any line parallel to $\sqrt{3}x + y + 3 = 0$ is of the form $\sqrt{3}x + y + k = 0$. Since this line is a tangent to the circle $x^2 + y^2 = a^2$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $a$. Using the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,which is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$,we get: $\frac{|\sqrt{3}(0) + 1(0) + k|}{\sqrt{(\sqrt{3})^2 + 1^2}} = a$ $\frac{|k|}{\sqrt{3 + 1}} = a$ $\frac{|k|}{2} = a$ $|k| = 2a$ $k = \pm 2a$ Substituting the value of $k$ back into the equation,we get $\sqrt{3}x + y \pm 2a = 0$.

The equations of the tangents to the circle $x^2 + y^2 = a^2$ parallel to the line $\sqrt{3}x + y + 3 = 0$ are

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