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Q)

# If the line $y=\sqrt 3x+k$ touches the circle $x^2+y^2=16$,then find the value of $k$.

[Hint: Equate perpendicular distance from the centre of the circle to its radius].

$\begin{array}{1 1}k=8\\k=\pm 9\\k=\pm 8\\k=-9\\\end{array}$

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A)
Toolbox:
• The radius of a circle is always perpendicular to the tangent.
• The perpendicular distance from a point $(x_1,y_1)$ to the line $ax+by+c=0$ is $d=\large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$
Answer : $k=\pm 8$
Equation of the line which touches the circle is $y=\sqrt 3x+k\Rightarrow \sqrt 3x-y+k=0$
Equation of the circle is $x^2+y^2=16$
$\Rightarrow$ The coordinate of the origin is $(0,0)$, radius $'r'=4$.
The length of the perpendicular from $(0,0)$ to the line is the radius.
$\therefore 4=\bigg|\large\frac{\sqrt 3(0)-(0)+k}{\sqrt{(\sqrt 3)^2+(1)^2}}\bigg|$
$\Rightarrow 4 = \large\frac{k}{2}$
$\therefore k =\pm 8$