Question

If the line $y=\sqrt{3}x+k$ touches the circle x² + y² = 16, then k = __________.

Answer 1

$\mathrm{Given}\mathrm{line}y=\sqrt{3}x+k\mathrm{and}\mathrm{the}\mathrm{circle}{x}^2+y^2=16$

i.e. Center of the circle is (0, 0) and radius is 4

Since line $y=\sqrt{3}x+k$ touches the circle, perpendicular distance from (0, 0) to the line is equal to the radius of the circle

$$\begin{gathered} \mathrm{Hence},4=\left|\frac{0-0+k}{\sqrt{3+1}}\right| \\ \mathrm{i}.\mathrm{e}.4=\pm \frac{k}{2} \\ \mathrm{i}.\mathrm{e}.k=\pm 8 \end{gathered}$$