Question

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

Answer 1

The given circle is $x^2+y^2=16$ ..........$\left(1\right)$

Its centre is $C(0,0)$ and its radius is $4$

The given line is $x-\sqrt{3}y-k=0$ ..........$\left(2\right)$

The line $\left(2\right)$ will touch the circle $\left(1\right)$ if the perpendicular distance from $C(0,0)$ to the line $\left(2\right)=$radius of circle.

$\Rightarrow \frac{|1\times 0-\sqrt{3}\times 0-k|}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}=4$

$\Rightarrow \frac{\left|k\right|}{2}=4$

$\Rightarrow k=\pm 8$