Condition of Tangency to Circle whose Center is Origin

Theorem
Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.

Let $\LL$ be a straight line in the plane of $\CC$ whose equation is given by:
 * $(1): \quad l x + m y + n = 0$

such that $l \ne 0$.

Then $\LL$ is tangent to $\CC$ :
 * $\paren {l^2 + m^2} r^2 = n^2$

Proof
From Equation of Circle center Origin, $\CC$ can be described as:
 * $(2): \quad x^2 + y^2 = r^2$

Let $\LL$ intersect with $\CC$.

To find where this happens, we find $x$ and $y$ which satisfy both $(1)$ and $(2)$.

So:

This is a quadratic in $y$.

This corresponds to the two points of intersection of $\LL$ with $\CC$.

When $\LL$ is tangent to $\CC$, these two points coincide.

Hence $(3)$ has equal roots.

From Solution to Quadratic Equation, this happens when the discriminant of $(3)$ is zero.

That is: