Condition of Tangency to Circle whose Center is Origin

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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 $\CC$ if and only if:

$\paren {l^2 + m^2} r^2 = n^2$


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:

\(\text {(1)}: \quad\) \(\ds l x + m y + n\) \(=\) \(\ds 0\) Equation for $\LL$
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds -\dfrac {m y} l - \dfrac n l\) rearranging
\(\ds \leadsto \ \ \) \(\ds \paren {-\dfrac {m y} l - \dfrac n l}^2 + y^2\) \(=\) \(\ds r^2\) substituting for $x$ in $(2)$
\(\ds \leadsto \ \ \) \(\ds \paren {-m y - n}^2 + l^2 y^2\) \(=\) \(\ds l^2 r^2\) multiplying by $l^2$
\(\ds \leadsto \ \ \) \(\ds m^2 y^2 + 2 m n y + n^2 + l^2 y^2\) \(=\) \(\ds l^2 r^2\) multiplying out
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \paren {l^2 + m^2} y^2 + 2 m n y + \paren {n^2 - l^2 r^2}\) \(=\) \(\ds 0\) rearranging

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:

\(\ds m^2 n^2\) \(=\) \(\ds \paren {l^2 + m^2} \paren {n^2 - l^2 r^2}\)
\(\ds \leadsto \ \ \) \(\ds l^2 n^2 - l^2 m^2 r^2 - l^4 r^2\) \(=\) \(\ds 0\) multiplying out and simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {l^2 + m^2} r^2\) \(=\) \(\ds n^2\) as $l^2 \ne 0$

$\blacksquare$


Sources

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