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If the line $y=\sqrt{3} x+k$ touches the circle $x^2+y^2=16$, then find the value of $k$.
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Perpendicular distance from centre $(0,0)$ on line $(y=\sqrt{3 x}+k)$ is the radius of the given circle.
$\begin{array}{lr}\Rightarrow \frac{|k|}{\sqrt{3+1}}=4 \Rightarrow k=\pm 8 & \Rightarrow b^2=\frac{45}{4} \\ \text { Here; } 2 \mathrm{~g}=-6 \mathrm{~g} \& 2 \mathrm{f}=12 \Rightarrow \mathrm{g}=-3 \& \mathrm{f}=6 & \therefore \text { Equation of } \\ \frac{x^2}{(9)^2}+\frac{y^2}{45}\end{array}$
$\begin{array}{lr}\Rightarrow \frac{|k|}{\sqrt{3+1}}=4 \Rightarrow k=\pm 8 & \Rightarrow b^2=\frac{45}{4} \\ \text { Here; } 2 \mathrm{~g}=-6 \mathrm{~g} \& 2 \mathrm{f}=12 \Rightarrow \mathrm{g}=-3 \& \mathrm{f}=6 & \therefore \text { Equation of } \\ \frac{x^2}{(9)^2}+\frac{y^2}{45}\end{array}$
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