A tangent PT is drawn to the circle x^2 + y^2 | vedclass

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(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at $P(\sqrt{3}, 1)$ is given by $x x_1 + y y_1 = r^2$,which is $\sqrt{3}x + y = 4$. The

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$x - \sqrt{3}y = 1$

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(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at $P(\sqrt{3}, 1)$ is given by $x x_1 + y y_1 = r^2$,which is $\sqrt{3}x + y = 4$. The slope of this tangent $PT$ is $m_1 = -\sqrt{3}$. The line $L$ is perpendicular to $PT$,so its slope $m_2$ satisfies $m_1 	imes m_2 = -1$,giving $m_2 = \frac{1}{\sqrt{3}}$. Let the equation of line $L$ be $y = \frac{1}{\sqrt{3}}x + c$,or $x - \sqrt{3}y + \sqrt{3}c = 0$. This line is tangent to the circle $(x - 3)^2 + y^2 = 1$,which has center $(3, 0)$ and radius $r = 1$. The perpendicular distance from the center $(3, 0)$ to the line $x - \sqrt{3}y + \sqrt{3}c = 0$ must equal the radius $1$: $\frac{|3 - \sqrt{3}(0) + \sqrt{3}c|}{\sqrt{1^2 + (-\sqrt{3})^2}} = 1$ $\frac{|3 + \sqrt{3}c|}{2} = 1$ $|3 + \sqrt{3}c| = 2$ Case $1$: $3 + \sqrt{3}c = 2 \implies \sqrt{3}c = -1 \implies c = -\frac{1}{\sqrt{3}}$. The equation is $y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \implies x - \sqrt{3}y = 1$. Case $2$: $3 + \sqrt{3}c = -2 \implies \sqrt{3}c = -5 \implies c = -\frac{5}{\sqrt{3}}$. The equation is $y = \frac{1}{\sqrt{3}}x - \frac{5}{\sqrt{3}} \implies x - \sqrt{3}y = 5$. Comparing with the given options,$x - \sqrt{3}y = 1$ is option $A$.

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

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