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

Concept: Distance from centre of circle to tangent line equals radius.

The circle $x^2 + y^2 = 16$ has centre $(0,0)$ and radius $4$.

The line is $y = \sqrt{3}\,x + k$, or in standard form:

$\sqrt{3}\,x - y + k = 0$.

Distance from centre $(0,0)$ to this line is …

The problem asks: for what value(s) of $k$ does the line $y = \sqrt{3}\,x + k$ just touch the circle $x^2 + y^2 = 16$? "Touches" means the line is a tangent — it meets the circle at exactly one point.

The circle $x^2 + y^2 = 16$ has centre at $(0,0)$ and radius $r = 4$ (since $r^2 = 16$).

The key idea: a line is tangent to a circle if and only if the perpendicular distance from the centre of the circle to the line is exactly equal to the radius. Why? Because the shortest distance from the centre to the line is along the perpendicular; if that distance equals the radius, the line just grazes the circle at one point. If it were less, the line would cut through (two intersections); if more, it would miss entirely.

So we compute the perpendicular distance from $(0,0)$ to the line $y = \sqrt{3}\,x + k$, set it equal to $4$, and solve for $k$.

1. Rewrite the line in standard form. The line is $y = \sqrt{3}\,x + k$. Bring all terms to one side:

This is of the form $Ax + By + C = 0$ with $A = \sqrt{3}$, $B = -1$, $C = k$.

2. Recall the distance formula. The perpendicular distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is:

Here $(x_1, y_1) = (0,0)$, so the numerator becomes $|A\cdot 0 + B\cdot 0 + C| = |C| = |k|$.

3. Compute the denominator. $A^2 + B^2 = (\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$ …