Question

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

Answer 1

Consider the given equation of a circle as $x^2+y^2=16$
Centre is (0, 0) and radius =4 as shown figure.

AB be a line passing through centre of circle.
Tangent $y=\sqrt{3}x+k$ touches the circle at B(a,b)
$a^2+b^2=16$
AB is perpendicular to tangent.
Slope of $AB=-\frac{1}{\sqrt{3}}$
Equation of AB is
$y=-\frac{1}{\sqrt{3}}\times$ [AB passes through centre (0, 0)]
$b=-\frac{1}{\sqrt{3}}a\dots \left(2\right)$
Substituting (2) in (1), we get,
$a^2+\frac{1}{3}{a}^2=16\frac{4a^2}{3}=16a=\pm \sqrt{3}b=\pm 2$
B(a,b) is on $y=\left(2\sqrt{3}\right)+k\pm 2=\pm 6+kk=\pm 8$