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A tangent PT is drawn to the circle $x^2+y^2=4$ at the point $\mathrm{P}(\sqrt{3}, 1)$. If a straight line $\mathrm{L}$ which is perpendicular to $\mathrm{PT}$ is a tangent to the circle $(\mathrm{x}-3)^2+\mathrm{y}^2=1$, then a possible equation of $\mathrm{L}$ is
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Verified Answer
The correct answer is:
$x-\sqrt{3} y=1$
Equation of tangent to circle $\mathrm{T}=0$
$\Rightarrow$ T. $\sqrt{3} x+y=4$...(i)
$$
m_{\mathrm{T}}=-\sqrt{3}
$$
$\therefore$ Slope of line perpendicular to $\mathrm{PT}=\frac{1}{\sqrt{3}}$
$\because \mathrm{L}$ is tangent of $(x-3)^2+y^2=1$
i.e. $x^2+y^2-6 x+8=0$
Let equation of tangent be
$y=\frac{1}{\sqrt{3}} x+C \Rightarrow x-\sqrt{3} y+C \sqrt{3}=0$...(ii)
Length of perpendicular on (i) from centre $(3,0)=$ radius $=1$
$$
1=\left|\frac{3+C \sqrt{3}}{2}\right| \Rightarrow C \sqrt{3}=-1,-5
$$
$\Rightarrow \mathrm{C}=\frac{-1}{\sqrt{3}}$ or $\frac{-5}{\sqrt{3}}$
$\therefore$ Equation of L: $y=\frac{1}{\sqrt{3}} x-\frac{1}{\sqrt{3}}$
$$
\Rightarrow \sqrt{3} y=x-1
$$
$\Rightarrow$ T. $\sqrt{3} x+y=4$...(i)
$$
m_{\mathrm{T}}=-\sqrt{3}
$$
$\therefore$ Slope of line perpendicular to $\mathrm{PT}=\frac{1}{\sqrt{3}}$
$\because \mathrm{L}$ is tangent of $(x-3)^2+y^2=1$
i.e. $x^2+y^2-6 x+8=0$
Let equation of tangent be
$y=\frac{1}{\sqrt{3}} x+C \Rightarrow x-\sqrt{3} y+C \sqrt{3}=0$...(ii)
Length of perpendicular on (i) from centre $(3,0)=$ radius $=1$
$$
1=\left|\frac{3+C \sqrt{3}}{2}\right| \Rightarrow C \sqrt{3}=-1,-5
$$
$\Rightarrow \mathrm{C}=\frac{-1}{\sqrt{3}}$ or $\frac{-5}{\sqrt{3}}$
$\therefore$ Equation of L: $y=\frac{1}{\sqrt{3}} x-\frac{1}{\sqrt{3}}$
$$
\Rightarrow \sqrt{3} y=x-1
$$
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