A tangent PT is drawn to the circle x^2+y^2=4 | 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 $T=0$,which is $\sqrt{3}x+y=4$. The slope of this tangent $P

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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 $T=0$,which is $\sqrt{3}x+y=4$. The slope of this tangent $PT$ is $m_{PT} = -\sqrt{3}$. Since line $L$ is perpendicular to $PT$,the slope of $L$ is $m_L = \frac{1}{\sqrt{3}}$. Let the equation of line $L$ be $y = \frac{1}{\sqrt{3}}x + c$,which can be rewritten as $x - \sqrt{3}y + \sqrt{3}c = 0$. Since $L$ is a tangent to the circle $(x-3)^2+y^2=1$ with center $(3, 0)$ and radius $r=1$,the perpendicular distance from the center to the line must equal the radius: $\frac{|3 - \sqrt{3}(0) + \sqrt{3}c|}{\sqrt{1^2 + (-\sqrt{3})^2}} = 1$ $\frac{|3 + \sqrt{3}c|}{2} = 1$ $|3 + \sqrt{3}c| = 2$ Case $1$: $3 + \sqrt{3}c = 2$ $\Rightarrow \sqrt{3}c = -1$ $\Rightarrow c = -\frac{1}{\sqrt{3}}$. The equation becomes $y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \Rightarrow x - \sqrt{3}y = 1$. Case $2$: $3 + \sqrt{3}c = -2$ $\Rightarrow \sqrt{3}c = -5$ $\Rightarrow c = -\frac{5}{\sqrt{3}}$. The equation becomes $y = \frac{1}{\sqrt{3}}x - \frac{5}{\sqrt{3}} \Rightarrow x - \sqrt{3}y = 5$.

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

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